#set document(title: "5.6 Quadratic Equations In One Variable with Applications", author: "OpenStax / XYZ Homework") #set page(width: 8.5in, height: auto, margin: 1in) #import "@preview/cetz:0.5.2" #set text(font: ("STIX Two Text", "Libertinus Serif", "New Computer Modern"), size: 10.5pt, lang: "en") #show math.equation: set text(font: ("STIX Two Math", "New Computer Modern Math")) #set par(justify: true, leading: 0.62em, spacing: 0.9em) #set enum(spacing: 1.1em) // room between list items so tall inline fractions don't collide #set list(spacing: 1.1em) #set table(stroke: 0.5pt + rgb("#c7ccd3")) #let BLUE = rgb("#183B6F") // brand navy — section bars + example/solution labels (white on navy 11.09:1) #let ORANGE = rgb("#A94509") // brand primary-700 — AA-safe deep orange for TEXT (5.93:1 on white; raw brand #F37021 is 2.94:1 and must never carry text) #let RED = rgb("#DC2626") // brand error-600 #let GREEN = rgb("#059669") // brand success-600 (decoration only; small green text uses green-text #007942) #show heading.where(level: 1): it => block(width: 100%, above: 0pt, below: 16pt, fill: gradient.linear(BLUE, rgb("#2C5AA0")), inset: (x: 14pt, y: 12pt), radius: 3pt, text(fill: white, weight: "bold", size: 19pt, it.body)) #show heading.where(level: 2): it => block(width: 100%, above: 18pt, below: 10pt, fill: BLUE, inset: (x: 10pt, y: 6pt), radius: 2pt, text(fill: white, weight: "bold", size: 12pt, it.body)) #show heading.where(level: 3): it => text(fill: ORANGE, weight: "bold", size: 12.5pt, it.body) #show heading.where(level: 4): it => text(fill: BLUE, weight: "bold", size: 10.5pt, it.body) #let examplebox(label, title, body) = block(width: 100%, breakable: true, fill: rgb("#EFF1F5"), stroke: 0.5pt + rgb("#CFDDF0"), radius: 4pt, inset: 10pt, above: 12pt, below: 12pt)[ #block(below: 6pt)[#box(fill: BLUE, inset: (x: 6pt, y: 2pt), radius: 2pt, text(fill: white, weight: "bold", size: 8.5pt, label)) #h(0.4em) #strong[#title]] #body] // rail = decorative left rule (raw brand token); labelcolor = AA-safe label text shade #let notebox(label, rail, labelcolor, tint, body) = block(width: 100%, breakable: true, fill: tint, stroke: (left: 3pt + rail), inset: (left: 10pt, rest: 8pt), radius: (right: 4pt), above: 11pt, below: 11pt)[ #text(fill: labelcolor, weight: "bold", size: 7.5pt, tracking: 0.5pt)[#upper(label)] #linebreak() #body] #let solutionbox(body) = block(above: 4pt, below: 8pt)[ #text(fill: BLUE, weight: "bold", size: 8.5pt)[Solution] #linebreak() #body] #let figph(msg) = block(width: 100%, height: 60pt, fill: rgb("#f6f7f9"), stroke: (paint: rgb("#c7ccd3"), dash: "dashed"), radius: 4pt, inset: 10pt)[ #align(center + horizon, text(fill: rgb("#889"), style: "italic", size: 9pt, msg))] // Standardize inlined figure sizes: measure the natural CeTZ canvas, then scale to a // consistent envelope (aspect-aware; see build_typst.py FIG_* constants). Unlike the // print preamble, dimensions are FLOORED: in an editor a user can trim a figure to a // degenerate 1-D shape (a bare line), and w/h or tw/w would then divide by zero. #let _STD_W = 3.5 #let _WIDE_W = 5.6 #let _MAX_H = 3.4 #let _ASPECT_WIDE = 2.2 #let _UPSCALE_MAX = 1.15 #let stdfig(body) = context { let m = measure(body) let w = calc.max(m.width / 1in, 0.01) let h = calc.max(m.height / 1in, 0.01) let tw = if w / h > _ASPECT_WIDE { _WIDE_W } else { _STD_W } let s = calc.min(tw / w, _MAX_H / h, _UPSCALE_MAX) align(center, box(scale(x: s * 100%, y: s * 100%, reflow: true, body))) } #show figure: set block(breakable: false) #set figure(gap: 8pt) #show figure.caption: set text(size: 8.5pt, fill: rgb("#555")) == 5.6#h(0.6em)Quadratic Equations In One Variable with Applications #figure(figph[The Gateway Arch in St. Louis, Missouri], alt: "The Gateway Arch in St. Louis, Missouri", caption: [The Gateway Arch in St. Louis, Missouri]) === Learning Objectives After completing this section, you should be able to: + Multiply binomials. + Factor trinomials. + Solve quadratic equations by graphing. + Solve quadratic equations by factoring. + Solve quadratic equations using square root method. + Solve quadratic equations using the quadratic formula. + Solve real world applications modeled by quadratic equations. In this section, we will discuss quadratic equations. There are several real-world scenarios that can be represented by the graph of a quadratic equation. Think of the Gateway Arch in St. Louis, Missouri. Both ends of the arch are 630 feet apart and the arch is 630 feet tall. You can plot these points on a coordinate system and create a parabola to graph the quadratic equation. === Identify Polynomials, Monomials, Binomials and Trinomials You have learned that a term is a constant, or the product of a constant and one or more variables. When it is of the form #math.equation(block: false, alt: "a x to the power m")[$a x^(m)$], where #math.equation(block: false, alt: "a")[$a$] is a constant and #math.equation(block: false, alt: "x to the power m")[$x^(m)$] is a positive whole number, it is called a #strong[monomial]. Some examples of monomial are 8, #math.equation(block: false, alt: "minus 2 x squared")[$− 2 x^(2)$], #math.equation(block: false, alt: "4 y cubed")[$4 y^(3)$], and #math.equation(block: false, alt: "11 z")[$11 z$]. A monomial or two or more monomials combined by addition or subtraction is a #strong[polynomial]. Some examples include: #math.equation(block: false, alt: "b plus 11")[$b + 11$], #math.equation(block: false, alt: "4 y squared minus 7 y plus 2")[$4 y^(2) − 7 y + 2$], and #math.equation(block: false, alt: "4 x to the power 4 plus x cubed plus 8 x squared minus 9 x plus 1")[$4 x^(4) + x^(3) + 8 x^(2) − 9 x + 1$]. Some polynomials have special names, based on the number of terms. A monomial is a polynomial with exactly one term (examples: 14, #math.equation(block: false, alt: "8 y squared")[$8 y^(2)$], #math.equation(block: false, alt: "minus 9 x cubed y to the power 5")[$− 9 x^(3) y^(5)$], and #math.equation(block: false, alt: "minus 13")[$− 13$]). A #strong[binomial] has exactly two terms (examples: #math.equation(block: false, alt: "a plus 7")[$a + " " 7$], #math.equation(block: false, alt: "4 b minus 5")[$4 b − 5$], #math.equation(block: false, alt: "y squared minus 16")[$y^(2) − 16$], and #math.equation(block: false, alt: "3 x cubed minus 9 x squared")[$3 x^(3) − 9 x^(2)$]), and a trinomial has exactly three terms (examples: #math.equation(block: false, alt: "x squared minus 7 x plus 12")[$x^(2) − 7 x + 12$], #math.equation(block: false, alt: "9 y squared plus 2 y minus 8")[$9 y^(2) + 2 y − 8$], #math.equation(block: false, alt: "6 m to the power 4 minus m cubed plus 8 m")[$6 m^(4) − m^(3) + 8 m$], and #math.equation(block: false, alt: "x to the power 4 plus 3 x squared minus 1")[$x^(4) + 3 x^(2) − 1$]). Notice that every monomial, binomial, and trinomial is also a polynomial. They are just special members of the “family” of polynomials and so they have special names. We use the words monomial, binomial, and trinomial when referring to these special polynomials and just call all the rest polynomials. === Multiply Binomials Recall multiplying algebraic expressions from Algebraic Expressions. In this section, we will continue that work and multiply binomials as well. We can use an area model to do multiplication. #examplebox("Example 1")[Multiply Binomials][ Multiply #math.equation(block: false, alt: "open parenthesis x plus 2 close parenthesis open parenthesis x plus 3 close parenthesis")[$( x + 2 ) ( x + 3 )$]. #solutionbox[ #strong[Step 1:] Use the distributive property: #math.equation(block: true, alt: "open parenthesis x close parenthesis open parenthesis x close parenthesis plus open parenthesis x close parenthesis open parenthesis 3 close parenthesis plus open parenthesis 2 close parenthesis open parenthesis x close parenthesis plus open parenthesis 2 close parenthesis open parenthesis 3 close parenthesis")[$( x ) ( x ) + ( x ) ( 3 ) + ( 2 ) ( x ) + ( 2 ) ( 3 )$] In the area model multiply each term on the side by each term on the top (think of it as a multiplication table). #figure( stdfig([#cetz.canvas({ import cetz.draw: * content((100.00pt, 197.33pt), text(font: sys.inputs.at("title-font", default: "STIX Two Text"), size: 11pt)[Area Diagram]) line((33.33pt, 33.33pt), (166.67pt, 33.33pt), (166.67pt, 133.33pt), (33.33pt, 133.33pt), close: true, fill: none, stroke: 0.7pt + black) line((100.00pt, 33.33pt), (100.00pt, 133.33pt), stroke: 0.7pt + black) line((33.33pt, 83.33pt), (166.67pt, 83.33pt), stroke: 0.7pt + black) content((66.67pt, 108.33pt), anchor: "center", box(width: 58.67pt)[#align(center)[#text(size: 9pt)[x²]]]) content((133.33pt, 108.33pt), anchor: "center", box(width: 58.67pt)[#align(center)[#text(size: 9pt)[2x]]]) content((66.67pt, 58.33pt), anchor: "center", box(width: 58.67pt)[#align(center)[#text(size: 9pt)[3x]]]) content((133.33pt, 58.33pt), anchor: "center", box(width: 58.67pt)[#align(center)[#text(size: 9pt)[6]]]) content((16.67pt, 83.33pt), anchor: "center", text(size: 9pt)[x - 3]) content((100.00pt, 150.00pt), anchor: "center", text(size: 9pt)[x + 2]) })]), alt: "An area diagram shows a rectangle divided into two rows and two columns. The row headers read, x minus 3. The column headers read, x plus 2. The first row reads, x squared and 2 x. The second row reads, 3 x and 6.", caption: none, ) #strong[Step 2:] After we multiply, we get the following equation: #math.equation(block: true, alt: "x squared plus 3 x plus 2 x plus 6")[$x^(2) + 3 x + 2 x + 6$] #strong[Step 3:] Combine the like terms to arrive at: #math.equation(block: true, alt: "x squared plus 5 x plus 6")[$x^(2) + 5 x + 6$] ] ] #examplebox("Example 2")[Multiplying More Complex Binomials][ Multiply #math.equation(block: false, alt: "open parenthesis 2 x plus 7 close parenthesis open parenthesis 3 x minus 5 close parenthesis")[$( 2 x + 7 ) ( 3 x − 5 )$]. #solutionbox[ #strong[Step 1:] Use the Distributive Property: #math.equation(block: true, alt: "open parenthesis 2 x close parenthesis open parenthesis 3 x close parenthesis minus open parenthesis 2 x close parenthesis open parenthesis 5 close parenthesis plus open parenthesis 7 close parenthesis open parenthesis 3 x close parenthesis minus open parenthesis 7 close parenthesis open parenthesis 5 close parenthesis")[$( 2 x ) ( 3 x ) − ( 2 x ) ( 5 ) + ( 7 ) ( 3 x ) − ( 7 ) ( 5 )$] #figure( stdfig([#cetz.canvas({ import cetz.draw: * content((100.00pt, 174.00pt), text(font: sys.inputs.at("title-font", default: "STIX Two Text"), size: 11pt)[Area Diagram]) line((20.00pt, 20.00pt), (180.00pt, 20.00pt), stroke: 0.7pt + black) line((20.00pt, 140.00pt), (180.00pt, 140.00pt), stroke: 0.7pt + black) line((20.00pt, 20.00pt), (20.00pt, 140.00pt), stroke: 0.7pt + black) line((180.00pt, 20.00pt), (180.00pt, 140.00pt), stroke: 0.7pt + black) line((100.00pt, 20.00pt), (100.00pt, 140.00pt), stroke: 0.7pt + black) line((20.00pt, 80.00pt), (180.00pt, 80.00pt), stroke: 0.7pt + black) content((60.00pt, 150.00pt), anchor: "center", box(width: 72.00pt)[#align(center)[#text(size: 9pt)[2x]]]) content((140.00pt, 150.00pt), anchor: "center", box(width: 72.00pt)[#align(center)[#text(size: 9pt)[7]]]) content((10.00pt, 110.00pt), anchor: "center", box(width: 16.00pt)[#align(center)[#text(size: 9pt)[3x]]]) content((10.00pt, 50.00pt), anchor: "center", box(width: 16.00pt)[#align(center)[#text(size: 9pt)[-5]]]) content((60.00pt, 110.00pt), anchor: "center", box(width: 42.00pt)[#align(center)[#text(size: 9pt)[6x²]]]) content((140.00pt, 110.00pt), anchor: "center", box(width: 72.00pt)[#align(center)[#text(size: 9pt)[21x]]]) content((60.00pt, 50.00pt), anchor: "center", box(width: 42.00pt)[#align(center)[#text(size: 9pt)[-10x]]]) content((140.00pt, 50.00pt), anchor: "center", box(width: 72.00pt)[#align(center)[#text(size: 9pt)[-35]]]) })]), alt: "An area diagram shows a rectangle divided into two rows and two columns. The row headers read, 3 x minus 5. The column headers read, 2 x plus 7. The first row reads, 6 x squared and 21 x. The second row reads, negative 10 x and negative 35.", caption: none, ) #strong[Step 2:] After multiplying, get the following equation: #math.equation(block: true, alt: "6 x squared minus 10 x plus 21 x minus 35")[$6 x^(2) − 10 x + 21 x − 35$] #strong[Step 3:] Combine the like terms to arrive at: #math.equation(block: true, alt: "6 x squared plus 11 x minus 35")[$6 x^(2) + 11 x − 35$] ] ] #notebox("Who Knew?", rgb("#183B6F"), rgb("#183B6F"), rgb("#EFF1F5"))[ #emph[They Are Teaching Multiplication of Binomials in Elementary School] Manipulatives are often used in elementary school for students to experience a hands-on way to experience the mathematics they are learning. Base Ten Blocks, or Dienes Blocks, are often used to introduce place value and the operation of whole numbers. When multiplying two-digit numbers, students can make an array to visualize the Distributive Property. shows the value of each Base Ten Block and shows how to multiply 17 and 23 using an area model and Base Ten Blocks. You can see how this helps students visualize the multiplication using the Distributive Property. Consider how #math.equation(block: false, alt: "open parenthesis 10 plus 7 close parenthesis open parenthesis 20 plus 3 close parenthesis")[$( 10 + 7 ) ( 20 + 3 )$] can easily extend to #math.equation(block: false, alt: "open parenthesis 10 plus x close parenthesis open parenthesis 20 plus y close parenthesis")[$( 10 + x ) ( 20 + y )$] in algebra! #figure(figph[Three Base 10 blocks are shown. The blocks have the following values: 100, 10, and 1.], alt: "Three Base 10 blocks are shown. The blocks have the following values: 100, 10, and 1.", caption: [The Value of Each Base Ten Block]) #figure( stdfig([#cetz.canvas({ import cetz.draw: * content((100.00pt, 183.23pt), text(font: sys.inputs.at("title-font", default: "STIX Two Text"), size: 11pt)[Area Model: 17 × 23 = 391]) line((15.38pt, 23.08pt), (15.38pt, 153.85pt), stroke: 0.7pt + black) line((92.31pt, 23.08pt), (92.31pt, 153.85pt), stroke: 0.7pt + black) line((169.23pt, 23.08pt), (169.23pt, 153.85pt), stroke: 0.7pt + black) line((192.31pt, 23.08pt), (192.31pt, 153.85pt), stroke: 0.7pt + black) line((15.38pt, 23.08pt), (192.31pt, 23.08pt), stroke: 0.7pt + black) line((15.38pt, 100.00pt), (192.31pt, 100.00pt), stroke: 0.7pt + black) line((15.38pt, 107.69pt), (192.31pt, 107.69pt), stroke: 0.7pt + black) line((15.38pt, 115.38pt), (192.31pt, 115.38pt), stroke: 0.7pt + black) line((15.38pt, 123.08pt), (192.31pt, 123.08pt), stroke: 0.7pt + black) line((15.38pt, 130.77pt), (192.31pt, 130.77pt), stroke: 0.7pt + black) line((15.38pt, 138.46pt), (192.31pt, 138.46pt), stroke: 0.7pt + black) line((15.38pt, 146.15pt), (192.31pt, 146.15pt), stroke: 0.7pt + black) line((15.38pt, 153.85pt), (192.31pt, 153.85pt), stroke: 0.7pt + black) content((53.85pt, 61.54pt), anchor: "center", box(width: 68.92pt)[#align(center)[#text(size: 9pt)[100]]]) content((130.77pt, 61.54pt), anchor: "center", box(width: 42.00pt)[#align(center)[#text(size: 9pt)[100]]]) content((180.77pt, 61.54pt), anchor: "center", box(width: 30.46pt)[#align(center)[#text(size: 9pt)[30]]]) content((53.85pt, 103.85pt), anchor: "center", box(width: 68.92pt)[#align(center)[#text(size: 9pt)[10]]]) content((130.77pt, 103.85pt), anchor: "center", box(width: 42.00pt)[#align(center)[#text(size: 9pt)[10]]]) content((180.77pt, 103.85pt), anchor: "center", box(width: 30.46pt)[#align(center)[#text(size: 9pt)[3]]]) content((53.85pt, 111.54pt), anchor: "center", box(width: 68.92pt)[#align(center)[#text(size: 9pt)[10]]]) content((130.77pt, 111.54pt), anchor: "center", box(width: 42.00pt)[#align(center)[#text(size: 9pt)[10]]]) content((180.77pt, 111.54pt), anchor: "center", box(width: 30.46pt)[#align(center)[#text(size: 9pt)[3]]]) content((53.85pt, 119.23pt), anchor: "center", box(width: 68.92pt)[#align(center)[#text(size: 9pt)[10]]]) content((130.77pt, 119.23pt), anchor: "center", box(width: 42.00pt)[#align(center)[#text(size: 9pt)[10]]]) content((180.77pt, 119.23pt), anchor: "center", box(width: 30.46pt)[#align(center)[#text(size: 9pt)[3]]]) content((53.85pt, 126.92pt), anchor: "center", box(width: 68.92pt)[#align(center)[#text(size: 9pt)[10]]]) content((130.77pt, 126.92pt), anchor: "center", box(width: 42.00pt)[#align(center)[#text(size: 9pt)[10]]]) content((180.77pt, 126.92pt), anchor: "center", box(width: 30.46pt)[#align(center)[#text(size: 9pt)[3]]]) content((53.85pt, 134.62pt), anchor: "center", box(width: 68.92pt)[#align(center)[#text(size: 9pt)[10]]]) content((130.77pt, 134.62pt), anchor: "center", box(width: 42.00pt)[#align(center)[#text(size: 9pt)[10]]]) content((180.77pt, 134.62pt), anchor: "center", box(width: 30.46pt)[#align(center)[#text(size: 9pt)[3]]]) content((53.85pt, 142.31pt), anchor: "center", box(width: 68.92pt)[#align(center)[#text(size: 9pt)[10]]]) content((130.77pt, 142.31pt), anchor: "center", box(width: 42.00pt)[#align(center)[#text(size: 9pt)[10]]]) content((180.77pt, 142.31pt), anchor: "center", box(width: 30.46pt)[#align(center)[#text(size: 9pt)[3]]]) content((53.85pt, 150.00pt), anchor: "center", box(width: 68.92pt)[#align(center)[#text(size: 9pt)[10]]]) content((130.77pt, 150.00pt), anchor: "center", box(width: 42.00pt)[#align(center)[#text(size: 9pt)[10]]]) content((180.77pt, 150.00pt), anchor: "center", box(width: 30.46pt)[#align(center)[#text(size: 9pt)[3]]]) content((11.54pt, 61.54pt), anchor: "east", text(size: 9pt)[10]) content((11.54pt, 103.85pt), anchor: "east", text(size: 9pt)[1]) content((11.54pt, 111.54pt), anchor: "east", text(size: 9pt)[1]) content((11.54pt, 119.23pt), anchor: "east", text(size: 9pt)[1]) content((11.54pt, 126.92pt), anchor: "east", text(size: 9pt)[1]) content((11.54pt, 134.62pt), anchor: "east", text(size: 9pt)[1]) content((11.54pt, 142.31pt), anchor: "east", text(size: 9pt)[1]) content((11.54pt, 150.00pt), anchor: "east", text(size: 9pt)[1]) content((53.85pt, 157.69pt), anchor: "south", text(size: 9pt)[10]) content((130.77pt, 157.69pt), anchor: "south", text(size: 9pt)[10]) content((180.77pt, 157.69pt), anchor: "south", text(size: 9pt)[3]) content((103.85pt, 11.54pt), anchor: "north", text(size: 9pt)[17 × 23 = 391]) })]), alt: "An area model shows the equation 17 times 23 equals 391. The number 17 is represented down the left side of the model as one 10 block and seven 1 blocks. The number 23 is represented along the top of the model as two 10 blocks and three 1 blocks. Inside the model, the first row shows two 100 blocks and three 10 blocks. The second row shows two 10 blocks and three 1 blocks. The third row shows two 10 blocks and three 1 blocks. The fourth row shows two 10 blocks and three 1 blocks. The fifth row shows two 10 blocks and three 1 blocks. The sixth row shows two 10 blocks and three 1 blocks. The seventh row shows two 10 blocks and three 1 blocks. The eighth row shows two 10 blocks and three 1 blocks.", caption: [How to Multiply 17 and 23 Using an Area Model and Base Ten Blocks], ) ] === Factoring Trinomials We’ve just covered how to multiply binomials. Now you will need to “undo” this multiplication—to start with the product and end up with the factors. Let us review an example of multiplying binomials to refresh your memory. #math.equation(block: true, alt: "open parenthesis x plus 2 close parenthesis open parenthesis x plus 3 close parenthesis equals x squared plus 5 x plus 6")[$( x + 2 ) ( x + 3 ) = x^(2) + 5 x + 6$] To factor the trinomial means to start with the product, #math.equation(block: false, alt: "x squared plus 5 x plus 6")[$x^(2) + 5 x + 6$], and end with the factors, #math.equation(block: false, alt: "open parenthesis x plus 2 close parenthesis open parenthesis x plus 3 close parenthesis")[$( x + 2 ) ( x + 3 )$]. You need to think about where each of the terms in the trinomial came from. The first term came from multiplying the first term in each binomial. So, to get #math.equation(block: false, alt: "x squared")[$x^(2)$] in the product, each binomial must start with an #math.equation(block: false, alt: "x")[$x$]. #math.equation(block: true, alt: "x squared plus 5 x plus 6; open parenthesis x close parenthesis open parenthesis x close parenthesis")[$x^(2) + 5 x + 6 \ ( italic(x) ) ( italic(x) )$] The last term in the trinomial came from multiplying the last term in each binomial. So, the last terms must multiply to 6. What two numbers multiply to 6? The factors of 6 could be 1 and 6, or 2 and 3. How do you know which pair to use? Consider the middle term. It came from adding the outer and inner terms. So the numbers that must have a product of 6 will need a sum of 5. We’ll test both possibilities and summarize the results in the following table, which will be very helpful when you work with numbers that can be factored in many different ways. #figure(table( columns: 2, align: left, inset: 6pt, table.header([Factors of 6], [Sum of Factors]), [1, 6], [#math.equation(block: false, alt: "1 plus 6 equals 7")[$1 + 6 = 7$]], [2, 3], [#math.equation(block: false, alt: "2 plus 3 equals 5")[$2 + 3 = 5$]], )) We see that 2 and 3 are the numbers that multiply to 6 and add to 5. We have the factors of #math.equation(block: false, alt: "x squared plus 5 x plus 6")[$x^(2) + 5 x + 6$]. They are #math.equation(block: false, alt: "open parenthesis x plus 2 close parenthesis open parenthesis x plus 3 close parenthesis")[$( x + 2 ) ( x + 3 )$]. #math.equation(block: true, alt: "x squared plus 5 x plus 6, product; open parenthesis x plus 2 close parenthesis open parenthesis x plus 3 close parenthesis, factors")[$x^(2) + 5 x + 6 & "product" \ ( x + 2 ) ( x + 3 ) & "factors"$] You can check if the factors are correct by multiplying. Looking back, we started with #math.equation(block: false, alt: "x squared plus 5 x plus 6")[$x^(2) + 5 x + 6$], which is of the form #math.equation(block: false, alt: "x squared plus b x plus c")[$x^(2) + b x + c$], where #math.equation(block: false, alt: "b equals 5")[$b = 5$] and #math.equation(block: false, alt: "c equals 6")[$c = 6$]. We factored it into two binomials of the form #math.equation(block: false, alt: "open parenthesis x plus m close parenthesis")[$( x + m )$] and #math.equation(block: false, alt: "open parenthesis x plus n close parenthesis")[$( x + n )$]. #math.equation(block: true, alt: "x squared plus 5 x plus 6, x squared plus b x plus c; open parenthesis x plus 2 close parenthesis open parenthesis x plus 3 close parenthesis, open parenthesis x plus m close parenthesis open parenthesis x plus n close parenthesis")[$x^(2) + 5 x + 6 & x^(2) + b x + c \ ( x + 2 ) ( x + 3 ) & ( x + m ) ( x + n )$] To get the correct factors, we found two number #math.equation(block: false, alt: "m")[$m$] and #math.equation(block: false, alt: "n")[$n$] whose product is #math.equation(block: false, alt: "c")[$c$] and sum is #math.equation(block: false, alt: "b")[$b$]. With the area model , start with an empty box and then put in the #math.equation(block: false, alt: "x squared")[$x^(2)$] term and #math.equation(block: false, alt: "c")[$c$]. #figure( stdfig([#cetz.canvas({ import cetz.draw: * line((33.33pt, 33.33pt), (166.67pt, 33.33pt), (166.67pt, 166.67pt), (33.33pt, 166.67pt), close: true, fill: none, stroke: 0.7pt + black) line((100.00pt, 33.33pt), (100.00pt, 166.67pt), stroke: 0.7pt + black) line((33.33pt, 100.00pt), (166.67pt, 100.00pt), stroke: 0.7pt + black) content((66.67pt, 133.33pt), anchor: "center", text(size: 9pt)[x²]) content((133.33pt, 66.67pt), anchor: "center", text(size: 9pt)[6]) })]), alt: "An area diagram shows a rectangle divided into two rows and two columns. The first row reads, x squared and nil. The second row reads, nil and 6.", caption: none, ) Continue by putting in two terms that add up to #math.equation(block: false, alt: "5 x : 2 x")[$5 x : 2 x$] and #math.equation(block: false, alt: "3 x")[$3 x$] : #figure( stdfig([#cetz.canvas({ import cetz.draw: * content((100.00pt, 147.33pt), text(font: sys.inputs.at("title-font", default: "STIX Two Text"), size: 11pt)[Area diagram]) line((0.00pt, 0.00pt), (200.00pt, 0.00pt), (200.00pt, 133.33pt), (0.00pt, 133.33pt), close: true, fill: none, stroke: 0.7pt + black) line((100.00pt, 0.00pt), (100.00pt, 133.33pt), stroke: 0.7pt + black) line((0.00pt, 66.67pt), (200.00pt, 66.67pt), stroke: 0.7pt + black) content((50.00pt, 100.00pt), anchor: "center", box(width: 92.00pt)[#align(center)[#text(size: 9pt)[x²]]]) content((150.00pt, 100.00pt), anchor: "center", box(width: 92.00pt)[#align(center)[#text(size: 9pt)[2x]]]) content((50.00pt, 33.33pt), anchor: "center", box(width: 92.00pt)[#align(center)[#text(size: 9pt)[3x]]]) content((150.00pt, 33.33pt), anchor: "center", box(width: 92.00pt)[#align(center)[#text(size: 9pt)[6]]]) })]), alt: "An area diagram shows a rectangle divided into two rows and two columns. The first row reads, x squared and 2 x. The second row reads, 3 x and 6.", caption: none, ) Then you find the terms of the binomials on the top and side : #figure( stdfig([#cetz.canvas({ import cetz.draw: * line((33.33pt, 33.33pt), (166.67pt, 33.33pt), (166.67pt, 166.67pt), (33.33pt, 166.67pt), close: true, fill: none, stroke: 0.7pt + black) line((100.00pt, 33.33pt), (100.00pt, 166.67pt), stroke: 0.7pt + black) line((33.33pt, 100.00pt), (166.67pt, 100.00pt), stroke: 0.7pt + black) content((66.67pt, 133.33pt), anchor: "center", box(width: 58.67pt)[#align(center)[#text(size: 9pt)[x²]]]) content((133.33pt, 133.33pt), anchor: "center", box(width: 58.67pt)[#align(center)[#text(size: 9pt)[2x]]]) content((66.67pt, 66.67pt), anchor: "center", box(width: 58.67pt)[#align(center)[#text(size: 9pt)[3x]]]) content((133.33pt, 66.67pt), anchor: "center", box(width: 58.67pt)[#align(center)[#text(size: 9pt)[6]]]) content((23.33pt, 133.33pt), anchor: "east", text(size: 9pt)[x]) content((23.33pt, 66.67pt), anchor: "east", text(size: 9pt)[3]) content((66.67pt, 176.67pt), anchor: "south", text(size: 9pt)[x]) content((133.33pt, 176.67pt), anchor: "south", text(size: 9pt)[2]) })]), alt: "An area diagram shows a rectangle divided into two rows and two columns. The row headers read, x plus 3. The column headers read, x plus 2. The first row reads, x squared and 2 x. The second row reads, 3 x and 6.", caption: none, ) #examplebox("Example 3")[Factoring Trinomials][ Factor #math.equation(block: false, alt: "x squared plus 7 x plus 12")[$x^(2) + 7 x + 12$]. #solutionbox[ The numbers that must have a product of 12 will need a sum of 7. We will summarize the results in a table below. #figure(table( columns: 2, align: left, inset: 6pt, table.header([Factors of 12], [Sum of Factors]), [1, 12], [#math.equation(block: false, alt: "1 plus 12 equals 13")[$1 + 12 = 13$]], [2, 6], [#math.equation(block: false, alt: "2 plus 6 equals 8")[$2 + 6 = 8$]], [3, 4], [#math.equation(block: false, alt: "3 plus 4 equals 7")[$3 + 4 = 7$]], )) We see that 3 and 4 are the numbers that multiply to 12 and add to 7. The factors of #math.equation(block: false, alt: "x squared plus 7 x plus 12")[$x^(2) + 7 x + 12$] are #math.equation(block: false, alt: "open parenthesis x plus 3 close parenthesis open parenthesis x plus 4 close parenthesis")[$( x + 3 ) ( x + 4 )$]. ] ] #examplebox("Example 4")[Factoring More Complex Trinomials][ Factor #math.equation(block: false, alt: "x squared minus 11 x plus 28")[$x^(2) − 11 x + 28$]. #solutionbox[ The numbers that must have a product of 28 will need a sum of #math.equation(block: false, alt: "minus 11")[$− 11$]. We will summarize the results in a table. #figure(table( columns: 2, align: left, inset: 6pt, table.header([Factors of 28], [Sum of Factors]), [#math.equation(block: false, alt: "1 , 28")[$1 , 28$]], [#math.equation(block: false, alt: "1 plus 28 equals 29")[$1 + 28 = 29$]], [#math.equation(block: false, alt: "2 , 14")[$2 , 14$]], [#math.equation(block: false, alt: "2 plus 14 equals 16")[$2 + 14 = 16$]], [#math.equation(block: false, alt: "4 , 7")[$4 , 7$]], [#math.equation(block: false, alt: "4 plus 7 equals 11")[$4 + 7 = 11$]], )) We see that 4 and 7 are the numbers that multiply to 28 and add to 11. But we needed #math.equation(block: false, alt: "minus 11")[$− 11$], so we will need to use #math.equation(block: false, alt: "minus 4")[$− 4$] and #math.equation(block: false, alt: "minus 7")[$− 7$] because #math.equation(block: false, alt: "open parenthesis minus 4 close parenthesis open parenthesis minus 7 close parenthesis equals 28")[$( − 4 ) ( − 7 ) = 28$] and #math.equation(block: false, alt: "open parenthesis minus 4 close parenthesis plus open parenthesis minus 7 close parenthesis equals minus 11")[$( − 4 ) + ( − 7 ) = − 11$]. The factors of #math.equation(block: false, alt: "x squared minus 11 x plus 28")[$x^(2) − 11 x + 28$] are #math.equation(block: false, alt: "open parenthesis x minus 4 close parenthesis open parenthesis x minus 7 close parenthesis")[$( x − 4 ) ( x − 7 )$]. ] ] #notebox("Video", rgb("#DC2626"), rgb("#DC2626"), rgb("#f7f8fa"))[ #link("https://openstax.org/r/Factoring_with_the_Box")[Factoring with the Box Method (Area Model)] ] === Solving Quadratic Equations by Graphing We have already solved and graphed linear equations in Graphing Linear Equations and Inequalities, equations of the form #math.equation(block: false, alt: "A x plus B y equals C")[$A x + B y = C$]. In linear equations, the variables have no exponents. #strong[Quadratic equations] are equations in which the variable is squared. The following are some examples of quadratic equations: #math.equation(block: true, alt: "x squared plus 5 x plus 6 equals 0, 3 y squared plus 4 y equals 106, 4 u squared minus 81 equals 0, n open parenthesis n plus 1 close parenthesis equals 42")[$x^(2) + 5 x + 6 = 0 & 3 y^(2) + 4 y = 106 & 4 u^(2) − 81 = 0 & n ( n + 1 ) = 42$] The last equation does not appear to have the variable squared, but when we simplify the expression on the left, we will get #math.equation(block: false, alt: "n squared plus n")[$n^(2) + n$]. The general form of a quadratic equation is #math.equation(block: false, alt: "a x squared plus b x plus c equals 0")[$a x^(2) + b x + c = 0$], where #math.equation(block: false, alt: "a , b ,")[$a , b ,$] and #math.equation(block: false, alt: "c")[$c$] are real numbers, with #math.equation(block: false, alt: "a not equal to 0")[$a ≠ 0$]. Remember that a solution of an equation is a value of a variable that makes a true statement when substituted into the equation. The solutions of quadratic equations are the values of the variables that make the quadratic equation #math.equation(block: false, alt: "a x squared plus b x plus c equals 0")[$a x^(2) + b x + c = 0$] true. To solve quadratic equations, we need methods different than the ones we used in solving linear equations. We will start by solving a quadratic equation from its graph. Just like we started graphing linear equations by plotting points, we will do the same for quadratic equations. Let us look first at graphing the quadratic equation #math.equation(block: false, alt: "y equals x squared")[$y = x^(2)$]. We will choose integer values of #math.equation(block: false, alt: "x")[$x$] between #math.equation(block: false, alt: "minus 2")[$− 2$] and 2 and find their #math.equation(block: false, alt: "y")[$y$] values, as shown in the table below. #figure(table( columns: 2, align: left, inset: 6pt, [#math.equation(block: false, alt: "y equals x squared")[$y = x^(2)$]], [], [#math.equation(block: false, alt: "x")[$bold(italic(x))$]], [#math.equation(block: false, alt: "y")[$bold(italic(y))$]], [0], [0], [1], [1], [#math.equation(block: false, alt: "minus 1")[$− 1$]], [1], [2], [4], [#math.equation(block: false, alt: "minus 2")[$− 2$]], [4], )) Notice when we let #math.equation(block: false, alt: "x equals 1")[$x = 1$] and #math.equation(block: false, alt: "x equals minus 1")[$x = − 1$], we got the same value for #math.equation(block: false, alt: "y")[$y$]. #math.equation(block: true, alt: "y equals x squared, y equals 1 squared, y equals 1; y equals x squared, y equals open parenthesis minus 1 close parenthesis squared, y equals 1")[$y = x^(2) & y = 1^(2) & y = 1 \ y = x^(2) & y = attach(( − 1 ), t: 2) & y = 1$] The same thing happened when we let #math.equation(block: false, alt: "x equals 2")[$x = 2$] and #math.equation(block: false, alt: "x equals minus 2")[$x = − 2$]. Now, we will plot the points to show the graph of #math.equation(block: false, alt: "y equals x squared")[$y = x^(2)$]. #figure( stdfig([#cetz.canvas({ import cetz.draw: * merge-path(stroke: (paint: rgb("#D62128"), thickness: 0.9pt), fill: none, { line((68.05pt, 170.00pt), (68.28pt, 169.17pt)) line((68.28pt, 169.17pt), (69.48pt, 164.91pt)) line((69.48pt, 164.91pt), (70.68pt, 160.76pt)) line((70.68pt, 160.76pt), (71.88pt, 156.72pt)) line((71.88pt, 156.72pt), (73.07pt, 152.79pt)) line((73.07pt, 152.79pt), (74.27pt, 148.97pt)) line((74.27pt, 148.97pt), (75.47pt, 145.26pt)) line((75.47pt, 145.26pt), (76.67pt, 141.67pt)) line((76.67pt, 141.67pt), (77.86pt, 138.18pt)) line((77.86pt, 138.18pt), (79.06pt, 134.80pt)) line((79.06pt, 134.80pt), (80.26pt, 131.54pt)) line((80.26pt, 131.54pt), (81.46pt, 128.39pt)) line((81.46pt, 128.39pt), (82.66pt, 125.34pt)) line((82.66pt, 125.34pt), (83.85pt, 122.41pt)) line((83.85pt, 122.41pt), (85.05pt, 119.59pt)) line((85.05pt, 119.59pt), (86.25pt, 116.88pt)) 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169.17pt)) line((161.72pt, 169.17pt), (161.95pt, 170.00pt)) }) circle((76.67pt, 141.67pt), radius: 3.0pt, fill: black, stroke: none) circle((95.83pt, 99.17pt), radius: 3.0pt, fill: black, stroke: none) circle((115.00pt, 85.00pt), radius: 3.0pt, fill: black, stroke: none) circle((134.17pt, 99.17pt), radius: 3.0pt, fill: black, stroke: none) circle((153.33pt, 141.67pt), radius: 3.0pt, fill: black, stroke: none) line((0.00pt, 85.00pt), (234.00pt, 85.00pt), stroke: 0.7pt + black, mark: (end: ">", fill: black, size: 0.18)) line((115.00pt, 0.00pt), (115.00pt, 174.00pt), stroke: 0.7pt + black, mark: (end: ">", fill: black, size: 0.18)) line((0.00pt, 85.00pt), (0.00pt, 82.00pt), stroke: 0.5pt + black) content((0.00pt, 80.00pt), anchor: "north", text(size: 8pt)[−6]) line((19.17pt, 85.00pt), (19.17pt, 82.00pt), stroke: 0.5pt + black) content((19.17pt, 80.00pt), anchor: "north", text(size: 8pt)[−5]) line((38.33pt, 85.00pt), (38.33pt, 82.00pt), stroke: 0.5pt + black) content((38.33pt, 80.00pt), 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black) content((110.00pt, 70.83pt), anchor: "east", text(size: 8pt)[−1]) line((112.00pt, 99.17pt), (115.00pt, 99.17pt), stroke: 0.5pt + black) content((110.00pt, 99.17pt), anchor: "east", text(size: 8pt)[1]) line((112.00pt, 113.33pt), (115.00pt, 113.33pt), stroke: 0.5pt + black) content((110.00pt, 113.33pt), anchor: "east", text(size: 8pt)[2]) line((112.00pt, 127.50pt), (115.00pt, 127.50pt), stroke: 0.5pt + black) content((110.00pt, 127.50pt), anchor: "east", text(size: 8pt)[3]) line((112.00pt, 141.67pt), (115.00pt, 141.67pt), stroke: 0.5pt + black) content((110.00pt, 141.67pt), anchor: "east", text(size: 8pt)[4]) line((112.00pt, 155.83pt), (115.00pt, 155.83pt), stroke: 0.5pt + black) content((110.00pt, 155.83pt), anchor: "east", text(size: 8pt)[5]) line((112.00pt, 170.00pt), (115.00pt, 170.00pt), stroke: 0.5pt + black) content((110.00pt, 170.00pt), anchor: "east", text(size: 8pt)[6]) })]), alt: "A parabola is plotted on an x y coordinate plane. The x and y axes range from negative 6 to 6, in increments of 1. The parabola opens up and it passes through the following points, (negative 2, 4), (negative 1, 1), (0, 0), (1, 1), and (2, 4). Note: all values are approximate.", caption: none, ) The graph is not a line. This figure is called a parabola. Every quadratic equation has a graph that looks like this. When #math.equation(block: false, alt: "y equals 0")[$y = 0$] the solution to the quadratic #math.equation(block: false, alt: "y equals x squared")[$y = x^(2)$] is 0 because #math.equation(block: false, alt: "x squared equals 0")[$x^(2) = 0$] at #math.equation(block: false, alt: "x equals 0")[$x = 0$]. #examplebox("Example 5")[Graphing a Quadratic Equation][ Graph #math.equation(block: false, alt: "y equals x squared minus 1")[$y = x^(2) − 1$] and list the solutions to the quadratic equation. #solutionbox[ We will graph the equation by plotting points. #strong[Step 1:] Choose integer values for #math.equation(block: false, alt: "x")[$x$], substitute them into the equation, and solve for #math.equation(block: false, alt: "y")[$y$]. #strong[Step 2:] Record the values of the ordered pairs in the chart. #figure(table( columns: 2, align: left, inset: 6pt, [#math.equation(block: false, alt: "y equals x squared minus 1")[$y = x^(2) − 1$]], [], [#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "y")[$y$]], [0], [#math.equation(block: false, alt: "minus 1")[$− 1$]], [1], [0], [#math.equation(block: false, alt: "minus 1")[$− 1$]], [0], [2], [3], [#math.equation(block: false, alt: "minus 2")[$− 2$]], [3], )) #strong[Step 3:] Plot the points and then connect them with a smooth curve. The result will be the graph of the equation #math.equation(block: false, alt: "y equals x squared minus 1")[$y = x^(2) − 1$] . The solutions are #math.equation(block: false, alt: "x equals 1")[$x = 1$] and #math.equation(block: false, alt: "x equals minus 1")[$x = − 1$]. #figure( stdfig([#cetz.canvas({ import cetz.draw: * merge-path(stroke: (paint: rgb("#D62128"), thickness: 0.9pt), fill: none, { line((64.29pt, 170.00pt), (64.69pt, 168.45pt)) line((64.69pt, 168.45pt), (65.29pt, 166.14pt)) line((65.29pt, 166.14pt), (65.89pt, 163.86pt)) line((65.89pt, 163.86pt), (66.48pt, 161.60pt)) line((66.48pt, 161.60pt), (67.08pt, 159.38pt)) line((67.08pt, 159.38pt), (67.68pt, 157.18pt)) line((67.68pt, 157.18pt), (68.28pt, 155.00pt)) line((68.28pt, 155.00pt), (68.88pt, 152.86pt)) line((68.88pt, 152.86pt), (69.48pt, 150.74pt)) line((69.48pt, 150.74pt), (70.08pt, 148.65pt)) line((70.08pt, 148.65pt), (70.68pt, 146.59pt)) line((70.68pt, 146.59pt), (71.28pt, 144.56pt)) line((71.28pt, 144.56pt), (71.88pt, 142.55pt)) line((71.88pt, 142.55pt), (72.47pt, 140.57pt)) line((72.47pt, 140.57pt), (73.07pt, 138.62pt)) line((73.07pt, 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content((110.00pt, 155.83pt), anchor: "east", text(size: 8pt)[5]) line((112.00pt, 170.00pt), (115.00pt, 170.00pt), stroke: 0.5pt + black) content((110.00pt, 170.00pt), anchor: "east", text(size: 8pt)[6]) })]), alt: "A parabola is plotted on an x y coordinate plane. The x and y axes range from negative 6 to 6, in increments of 1. The parabola opens up and it passes through the following points, (negative 2, 3), (negative 1, 0), (0, negative 1), (1, 0), and (2, 3). Note: all values are approximate.", caption: none, ) ] ] #examplebox("Example 6")[Solving a Quadratic Equation From Its Graph][ Find the solutions of #math.equation(block: false, alt: "y equals x squared plus 5 x plus 4")[$y = x^(2) + 5 x + 4$] from its graph . #figure( stdfig([#cetz.canvas({ import cetz.draw: * merge-path(stroke: (paint: rgb("#D62128"), thickness: 0.9pt), fill: none, { line((46.00pt, 170.00pt), (46.10pt, 169.49pt)) line((46.10pt, 169.49pt), (46.59pt, 166.95pt)) line((46.59pt, 166.95pt), (47.09pt, 164.43pt)) line((47.09pt, 164.43pt), (47.59pt, 161.95pt)) line((47.59pt, 161.95pt), (48.08pt, 159.51pt)) line((48.08pt, 159.51pt), (48.58pt, 157.09pt)) line((48.58pt, 157.09pt), (49.07pt, 154.71pt)) line((49.07pt, 154.71pt), (49.57pt, 152.35pt)) line((49.57pt, 152.35pt), (50.06pt, 150.03pt)) line((50.06pt, 150.03pt), (50.56pt, 147.74pt)) line((50.56pt, 147.74pt), (51.06pt, 145.48pt)) line((51.06pt, 145.48pt), (51.55pt, 143.26pt)) line((51.55pt, 143.26pt), (52.05pt, 141.06pt)) 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(115.00pt, 170.00pt), stroke: 0.5pt + black) content((110.00pt, 170.00pt), anchor: "east", text(size: 8pt)[10]) })]), alt: "A parabola is plotted on an x y coordinate plane. The x and y axes range from negative 10 to 10, in increments of 1. The parabola opens up and it passes through the following points, (negative 5, 4), (negative 4, 0), (negative 2.5, negative 2.25), (negative 1, 0), and (0, 4). Note: all values are approximate.", caption: none, ) #solutionbox[ The solutions of a quadratic equations are the values of #math.equation(block: false, alt: "x")[$x$] that make the equation a true statement when set equal to zero (i.e. when #math.equation(block: false, alt: "y equals 0")[$y = 0$]). #math.equation(block: false, alt: "x squared plus 5 x plus 4 equals 0")[$x^(2) + 5 x + 4 = 0$] at #math.equation(block: false, alt: "x equals minus 4")[$x = − 4$] and #math.equation(block: false, alt: "x equals minus 1")[$x = − 1$]. ] ] === Solving Quadratic Equations by Factoring Another way of solving quadratic equations is by factoring. We will use the #strong[Zero Product Property] that says that if the product of two quantities is zero, it must be that at least one of the quantities is zero. The only way to get a product equal to zero is to multiply by zero itself. #examplebox("Example 7")[Solving a Quadratic Equation by Factoring][ Solve #math.equation(block: false, alt: "open parenthesis x plus 1 close parenthesis open parenthesis x minus 4 close parenthesis equals 0")[$( x + 1 ) ( x − 4 ) = 0$]. #solutionbox[ #figure(table( columns: 3, align: left, inset: 6pt, table.header([#strong[Step 1.] Set each factor equal to zero.], [The product equals zero, so at least one factor must equal zero.], [#math.equation(block: false, alt: "open parenthesis x plus 1 close parenthesis open parenthesis x minus 4 close parenthesis equals 0; x plus 1 equals 0 or x minus 4 equals 0")[$( x + 1 ) ( x − 4 ) = 0 \ x + 1 = 0 #h(1em) "or" #h(1em) x − 4 = 0$]]), [#strong[Step 2.] Solve the linear equations.], [Solve each equation.], [#math.equation(block: false, alt: "x equals minus 1 or x equals 4")[$x = − 1 #h(1em) "or" #h(1em) x = 4$]], [#strong[Step 3.] Check.], [Substitute each solution separately into the original equation.], [#math.equation(block: false, alt: "x, equals, minus 1; open parenthesis x plus 1 close parenthesis open parenthesis x minus 4 close parenthesis, equals, 0; open parenthesis minus 1 plus 1 close parenthesis open parenthesis minus 1 minus 4 close parenthesis, equals ?, 0; open parenthesis 0 close parenthesis open parenthesis minus 5 close parenthesis, equals ?, 0; 0, equals, 0 ✓")[$x & = & − 1 \ ( x + 1 ) ( x − 4 ) & = & 0 \ ( − 1 + 1 ) ( − 1 − 4 ) & limits(=)^(?) & 0 \ ( 0 ) ( − 5 ) & limits(=)^(?) & 0 \ 0 & = & 0 ✓$]], [], [], [#math.equation(block: false, alt: "x, equals, 4; open parenthesis x plus 1 close parenthesis open parenthesis x minus 4 close parenthesis, equals, 0; open parenthesis 4 plus 1 close parenthesis open parenthesis 4 minus 4 close parenthesis, equals ?, 0; open parenthesis 5 close parenthesis open parenthesis 0 close parenthesis, equals ?, 0; 0, equals, 0 ✓")[$x & = & 4 \ ( x + 1 ) ( x − 4 ) & = & 0 \ ( 4 + 1 ) ( 4 − 4 ) & limits(=)^(?) & 0 \ ( 5 ) ( 0 ) & limits(=)^(?) & 0 \ 0 & = & 0 ✓$]], )) ] ] #examplebox("Example 8")[Solve Another Quadratic Equation by Factoring][ Solve #math.equation(block: false, alt: "x squared plus 2 x minus 8 equals 0")[$x^(2) + 2 x − 8 = 0$]. #solutionbox[ #figure(table( columns: 3, align: left, inset: 6pt, table.header([#strong[Step 1.] Write the quadratic equation in standard form, #math.equation(block: false, alt: "ax squared plus bx plus c equals 0")[$italic("ax")^(2) + italic("bx") + c = 0$].], [The equation is already in standard form.], [#math.equation(block: false, alt: "x squared plus 2 x minus 8 equals 0")[$x^(2) + 2 x − 8 = 0$]]), [#strong[Step 2.] Factor the quadratic expression.], [Factor #math.equation(block: false, alt: "x squared plus 2 x minus 8; open parenthesis x plus 4 close parenthesis open parenthesis x minus 2 close parenthesis")[$x^(2) + 2 x − 8 \ ( x + 4 ) ( x − 2 )$]], [#math.equation(block: false, alt: "open parenthesis x plus 4 close parenthesis open parenthesis x minus 2 close parenthesis equals 0")[$( x + 4 ) ( x − 2 ) = 0$]], [#strong[Step 3.] Use the Zero Product Property.], [Set each factor equal to zero.], [#math.equation(block: false, alt: "x plus 4 equals 0 or x minus 2 equals 0")[$x + 4 = 0 #h(1em) "or" #h(1em) x − 2 = 0$]], [#strong[Step 4.] Solve the linear equations.], [We have two linear equations.], [#math.equation(block: false, alt: "x equals minus 4 or x equals 2")[$x = − 4 #h(1em) "or" #h(1em) x = 2$]], [#strong[Step 5.] Check.], [Substitute each solution separately into the original equation.], [#math.equation(block: false, alt: "x squared plus 2 x minus 8, equals, 0; x, equals, minus 4; open parenthesis minus 4 close parenthesis squared minus 2 open parenthesis minus 4 close parenthesis minus 8, equals ?, 0; 16 plus open parenthesis minus 8 close parenthesis minus 8, equals ?, 0; 0, equals, 0 ✓")[$x^(2) + 2 x − 8 & = & 0 \ x & = & − 4 \ attach(( − 4 ), t: 2) − 2 ( − 4 ) − 8 & limits(=)^(?) & 0 \ 16 + ( − 8 ) − 8 & limits(=)^(?) & 0 \ 0 & = & 0 ✓$]], [], [], [#math.equation(block: false, alt: "x squared plus 2 x minus 8, equals, 0; x, equals, 2; 2 squared minus 2 open parenthesis 2 close parenthesis minus 8, equals ?, 0; 4 plus 4 minus 8, equals ?, 0; 0, equals, 0 ✓")[$x^(2) + 2 x − 8 & = & 0 \ x & = & 2 \ 2^(2) − 2 ( 2 ) − 8 & limits(=)^(?) & 0 \ 4 + 4 − 8 & limits(=)^(?) & 0 \ 0 & = & 0 ✓$]], )) ] ] #notebox("Video", rgb("#DC2626"), rgb("#DC2626"), rgb("#f7f8fa"))[ #link("https://openstax.org/r/Zero_Property")[Solving Quadratics with the Zero Property] ] #notebox("Checkpoint", rgb("#059669"), rgb("#007942"), rgb("#EAF3EC"))[ #emph[Be careful to write the quadratic equation in standard form first. The equation must be set equal to zero in order for you to use the Zero Product Property! Often students start in Step 2 resulting in an incorrect solution. For example, #math.equation(block: false, alt: "x squared plus 2 x minus 15 equals minus 7")[$x^(2) + 2 x − 15 = − 7$] cannot be factored to #math.equation(block: false, alt: "open parenthesis x minus 3 close parenthesis open parenthesis x plus 5 close parenthesis equals minus 7")[$( x − 3 ) ( x + 5 ) = − 7$] and then solved by setting each factor equal to #math.equation(block: false, alt: "minus 7")[$− 7$].] #emph[The correct way to solve this quadratic equation is to set it equal to zero FIRST: #math.equation(block: false, alt: "x squared plus 2 x minus 15 plus 7 equals minus 7 plus 7")[$x^(2) + 2 x − 15 + 7 = − 7 + 7$] which becomes #math.equation(block: false, alt: "x squared plus 2 x minus 8 equals 0")[$x^(2) + 2 x − 8 = 0$], then continue to factor. See the table below for the correct way to apply the Zero Product Property.] #figure(table( columns: 5, align: left, inset: 6pt, [], [#math.equation(block: false, alt: "x squared plus 2 x minus 15 equals minus 7")[$x^(2) + 2 x − 15 = − 7$]], [#math.equation(block: false, alt: "x squared plus 2 x minus 15 equals minus 7")[$x^(2) + 2 x − 15 = − 7$]], [], [], [#strong[Step 1]], [Skipped], [#math.equation(block: false, alt: "x squared plus 2 x minus 15 plus 7 equals minus 7 plus 7; x squared plus 2 x minus 8 equals 0")[$x^(2) + 2 x − 15 + 7 = − 7 + 7 \ x^(2) + 2 x − 8 = 0$]], [], [], [#strong[Step 2]], [#math.equation(block: false, alt: "open parenthesis x minus 3 close parenthesis open parenthesis x plus 5 close parenthesis equals minus 7")[$( x − 3 ) ( x + 5 ) = − 7$]], [#math.equation(block: false, alt: "open parenthesis x minus 2 close parenthesis open parenthesis x plus 4 close parenthesis equals 0")[$( x − 2 ) ( x + 4 ) = 0$]], [], [], [#strong[Step 3]], [#math.equation(block: false, alt: "x minus 3 equals minus 7")[$x − 3 = − 7$]], [#math.equation(block: false, alt: "x plus 5 equals minus 7")[$x + 5 = − 7$]], [#math.equation(block: false, alt: "x minus 2 equals 0")[$x − 2 = 0$]], [#math.equation(block: false, alt: "x plus 4 equals 0")[$x + 4 = 0$]], [#strong[Step 4]], [#math.equation(block: false, alt: "x equals minus 4")[$x = − 4$]], [#math.equation(block: false, alt: "x equals minus 12")[$x = − 12$]], [#math.equation(block: false, alt: "x equals 2")[$x = 2$]], [#math.equation(block: false, alt: "x equals minus 4")[$x = − 4$]], [#strong[Step 5]], [#math.equation(block: false, alt: "open parenthesis minus 4 close parenthesis squared plus 2 open parenthesis minus 4 close parenthesis minus 15; equals 16 minus 8 minus 15; equals minus 7")[$attach(( − 4 ), t: 2) + 2 ( − 4 ) − 15 \ = 16 − 8 − 15 \ = − 7$]], [#math.equation(block: false, alt: "open parenthesis minus 12 close parenthesis squared plus 2 open parenthesis minus 12 close parenthesis minus 15; equals 144 minus 24 minus 1 equals 105; not equal to minus 7")[$attach(( − 12 ), t: 2) + 2 ( − 12 ) − 15 \ = 144 − 24 − 1 = 105 \ ≠ − 7$]], [#math.equation(block: false, alt: "open parenthesis 2 close parenthesis squared plus 2 open parenthesis 2 close parenthesis minus 8; equals 4 plus 4 minus 8; equals 0")[$attach(( 2 ), t: 2) + 2 ( 2 ) − 8 \ = 4 + 4 − 8 \ = 0$]], [#math.equation(block: false, alt: "open parenthesis minus 4 close parenthesis squared plus 2 open parenthesis minus 4 close parenthesis minus 18; equals 16 minus 8 minus 8; equals 0")[$attach(( − 4 ), t: 2) + 2 ( − 4 ) − 18 \ = 16 − 8 − 8 \ = 0$]], )) ] === Solving Quadratic Equations Using the Square Root Property We just solved some quadratic equations by factoring. Let us use factoring to solve the quadratic equation #math.equation(block: false, alt: "x squared equals 9")[$x^(2) = 9$]. #figure(table( columns: 3, align: left, inset: 6pt, table.header([#strong[Step 1:] Put the equation in standard form.], [#math.equation(block: false, alt: "x squared minus 9 equals 0")[$x^(2) − 9 = 0$]], []), [#strong[Step 2:] Factor the left side.], [#math.equation(block: false, alt: "open parenthesis x plus 3 close parenthesis open parenthesis x minus 3 close parenthesis equals 0")[$( x + 3 ) ( x − 3 ) = 0$]], [], [#strong[Step 3:] Use the Zero Product Property.], [#math.equation(block: false, alt: "x plus 3 equals 0")[$x + 3 " " = 0$]], [#math.equation(block: false, alt: "x minus 3 equals 0")[$x − 3 = 0$]], [#strong[Step 4:] Solve each equation.], [#math.equation(block: false, alt: "x equals minus 3")[$x = − 3$]], [#math.equation(block: false, alt: "x equals 3")[$x = 3$]], [#strong[Step 5:] Combine the two solutions into #math.equation(block: false, alt: "plus or minus")[$±$]], [#math.equation(block: false, alt: "x equals plus or minus 3")[$x = ± 3$]], [], )) The solution is read as “#math.equation(block: false, alt: "x")[$x$] is equal to positive or negative 3.” What happens when we have an equation like #math.equation(block: false, alt: "x squared equals 7")[$x^(2) = 7$]? Since 7 is not a perfect square, we cannot solve the equation by factoring. These equations are all of the form #math.equation(block: false, alt: "x squared equals k")[$x^(2) = k$]. We define the square root of a number in this way: If #math.equation(block: false, alt: "n squared equals m")[$n^(2) = m$], then #math.equation(block: false, alt: "n")[$n$] is a square root of #math.equation(block: false, alt: "m")[$m$]. This leads to the #strong[Square Root Property]. #examplebox("Example 9")[Using the Square Root Property to Solve a Quadratic Equation][ Solve using the square Root Property: #math.equation(block: false, alt: "x squared equals 169")[$x^(2) = 169$]. #solutionbox[ #figure(table( columns: 3, align: left, inset: 6pt, table.header([#strong[Step 1:] Use the Square Root Property.], [#math.equation(block: false, alt: "x equals plus or minus the square root of 169")[$x = ± sqrt(169)$]], []), [#strong[Step 2:] Simplify the radical.], [#math.equation(block: false, alt: "x equals plus or minus 13")[$x = ± 13$]], [], [#strong[Step 3:] Rewrite to show the two solutions.], [#math.equation(block: false, alt: "x equals 13")[$x = 13$]], [#math.equation(block: false, alt: "x equals minus 13")[$x = − 13$]], )) ] ] #examplebox("Example 10")[Using the Square Root Property to Solve Another Quadratic Equation][ Solve using the Square Root Property: #math.equation(block: false, alt: "144 q squared equals 25")[$144 q^(2) = 25$]. #solutionbox[ #strong[Step 1:] Solve for #math.equation(block: false, alt: "q")[$q$]. #math.equation(block: true, alt: "q squared equals the fraction 25 over 144")[$q^(2) = frac(25, 144)$] #strong[Step 2:] Use the Square Root Property. #math.equation(block: true, alt: "q equals plus or minus the square root of the fraction 25 over 144")[$q = ± sqrt(frac(25, 144))$] #strong[Step 3:] Simplify the radical. #math.equation(block: true, alt: "q equals plus or minus the fraction 5 over 12")[$q = ± frac(5, 12)$] #strong[Step 4:] Rewrite to show the two solutions. #linebreak() #math.equation(block: false, alt: "q equals the fraction 5 over 12")[$q = frac(5, 12)$], #math.equation(block: false, alt: "q equals minus the fraction 5 over 12")[$q = − frac(5, 12)$] ] ] === Solving Quadratic Equations Using the Quadratic Formula This last method we will look at for solving quadratic equations is the #strong[quadratic formula]. This method works for all quadratic equations, even the quadratic equations we could not factor! To use the quadratic formula, we substitute the values of #math.equation(block: false, alt: "a")[$a$], #math.equation(block: false, alt: "b")[$b$], and #math.equation(block: false, alt: "c")[$c$] into the expression on the right side of the formula. Then, we do all the math to simplify the expression. The result gives the solution(s) to the quadratic equation. #examplebox("Example 11")[Solving a Quadratic Equation Using the Quadratic Formula][ Solve using the quadratic formula: #math.equation(block: false, alt: "x squared minus 6 x plus 5 equals 0")[$x^(2) − 6 x + 5 = 0$]. #solutionbox[ #math.equation(block: true, alt: "x squared minus 6 x plus 5 equals 0")[$x^(2) − 6 x + 5 = 0$] This equation is in standard form. #math.equation(block: true, alt: "a x squared plus b x plus c, equals, 0; x squared minus 6 x plus 5, equals, 0")[$a x^(2) + b x + c & = & 0 \ x^(2) − 6 x + 5 & = & 0$] #strong[Step 1:] Identify the #math.equation(block: false, alt: "a")[$a$], #math.equation(block: false, alt: "b")[$b$], and #math.equation(block: false, alt: "c")[$c$] values. #math.equation(block: true, alt: "a equals 1 , b equals minus 6 , c equals 5")[$a = 1 , b = − 6 , c = 5$] #strong[Step 2:] Write the quadratic formula. #math.equation(block: true, alt: "x equals the fraction minus b plus or minus the square root of b squared minus 4 a c over 2 a")[$x = frac(− b ± sqrt(b^(2) − 4 a c), 2 a)$] #strong[Step 3:] Substitute in the values of #math.equation(block: false, alt: "a")[$a$], #math.equation(block: false, alt: "b")[$b$], #math.equation(block: false, alt: "c")[$c$]. #math.equation(block: true, alt: "x equals the fraction open parenthesis minus 6 close parenthesis plus or minus the square root of open parenthesis minus 6 close parenthesis squared minus 4 open parenthesis 1 close parenthesis open parenthesis 5 close parenthesis over 2.1")[$x = frac(( − 6 ) ± sqrt(attach(( − 6 ), t: 2) − 4 ( 1 ) ( 5 )), 2.1)$] #strong[Step 4:] Simplify. #math.equation(block: true, alt: "x equals the fraction 6 plus or minus the square root of 36 minus 20 over 2")[$x = frac(6 ± sqrt(36 − 20), 2)$] #math.equation(block: true, alt: "x equals the fraction 6 plus or minus the square root of 16 over 2")[$x = frac(6 ± sqrt(16), 2)$] #math.equation(block: true, alt: "x equals the fraction 6 plus or minus 4 over 2")[$x = frac(6 ± 4, 2)$] #strong[Step 5:] Rewrite to show two solutions. #math.equation(block: true, alt: "x equals the fraction 6 plus 4 over 2 , x equals the fraction 6 minus 4 over 2")[$x = frac(6 + 4, 2) , x = frac(6 − 4, 2)$] #strong[Step 6:] Simplify. #math.equation(block: true, alt: "x equals the fraction 10 over 2 , x equals the fraction 2 over 2")[$x = frac(10, 2) , x = frac(2, 2)$] #math.equation(block: true, alt: "x equals 5 , x equals 1")[$x = 5 , x = 1$] #strong[Step 7:] Check. #math.equation(block: true, alt: "x squared minus 6 x plus 5, equals, 0, x squared minus 6 x plus 5, equals, 0; 5 squared minus 6 times 5 plus 5, equals ?, 0, 1 squared minus 6 times 1 plus 5, equals ?, 0; 25 minus 30 plus 5, equals ?, 0, 1 minus 6 plus 5, equals ?, 0; 0, equals, 0 ✓, 0, equals, 0 ✓")[$x^(2) − 6 x + 5 & = & 0 & x^(2) − 6 x + 5 & = & 0 \ 5^(2) − 6 ⋅ 5 + 5 & limits(=)^(?) & 0 & 1^(2) − 6 ⋅ 1 + 5 & limits(=)^(?) & 0 \ 25 − 30 + 5 & limits(=)^(?) & 0 & 1 − 6 + 5 & limits(=)^(?) & 0 \ 0 & = & 0 ✓ & 0 & = & 0 ✓$] ] ] #examplebox("Example 12")[Solving Another Quadratic Equation Using the Quadratic Formula][ Solve using the quadratic formula: #math.equation(block: false, alt: "2 x squared plus 9 x minus 5 equals 0")[$2 x^(2) + 9 x − 5 = 0$]. #solutionbox[ #figure(table( columns: 3, align: left, inset: 6pt, table.header([#strong[Step 1.] Write the quadratic equation in standard form. Identify the #math.equation(block: false, alt: "a")[$a$], #math.equation(block: false, alt: "b")[$b$], #math.equation(block: false, alt: "c")[$c$] values.], [This equation is in standard form.], [#math.equation(block: false, alt: "ax squared plus bx plus c equals 0; 2 x squared plus 9 x minus 5 equals 0; a equals 2 , b equals 9 , c equals minus 5")[$bold(italic("ax"))^(2) + bold(italic("bx")) + bold(italic(c)) = 0 \ 2 x^(2) + 9 x − 5 = 0 \ a = 2 , #h(0.28em) b = 9 , #h(0.28em) c = − 5$]]), [#strong[Step 2.] Write the quadratic formula. Then substitute in the values of #math.equation(block: false, alt: "a")[$a$], #math.equation(block: false, alt: "b")[$b$], #math.equation(block: false, alt: "c")[$c$].], [Substitute in #math.equation(block: false, alt: "a equals 2")[$bold(italic(a)) = 2$], #math.equation(block: false, alt: "b equals 9")[$bold(italic(b)) = 9$], #math.equation(block: false, alt: "c equals minus 5")[$bold(italic(c)) = − 5$]], [#math.equation(block: false, alt: "x equals the fraction minus b plus or minus the square root of b squared minus 4 a c over 2 a; x equals the fraction minus 9 plus or minus the square root of 9 squared minus 4 open parenthesis 2 close parenthesis open parenthesis minus 5 close parenthesis over 2 times 2")[$x = frac(− b ± sqrt(b^(2) − 4 a c), 2 a) \ x = frac(− 9 ± sqrt(9^(2) − 4 ( 2 ) ( − 5 )), 2 ⋅ 2)$]], [#strong[Step 3.] Simplify the fraction, and solve for #math.equation(block: false, alt: "x")[$x$].], [], [#math.equation(block: false, alt: "x equals the fraction minus 9 plus or minus the square root of 81 minus open parenthesis minus 40 close parenthesis over 4; x equals the fraction minus 9 plus or minus the square root of 121 over 4; x equals the fraction minus 9 plus or minus 11 over 4; x equals the fraction minus 9 plus 11 over 4 x equals the fraction minus 9 minus 11 over 4; x equals the fraction 2 over 4 x equals the fraction minus 20 over 4; x equals the fraction 1 over 2 x equals minus 5")[$x = frac(− 9 ± sqrt(81 − ( − 40 )), 4) \ x = frac(− 9 ± sqrt(121), 4) \ x = frac(− 9 ± 11, 4) \ x = frac(− 9 + 11, 4) #h(1em) x = frac(− 9 − 11, 4) \ x = frac(2, 4) #h(2.9em) x = frac(− 20, 4) \ x = frac(1, 2) #h(2.9em) x = − 5$]], [#strong[Step 4.] Check the solutions.], [Put each answer in the original equation to check. Substitute #math.equation(block: false, alt: "x equals the fraction 1 over 2")[$x = frac(1, 2)$].], [#math.equation(block: false, alt: "2 x squared plus 9 x minus 5, equals, 0; 2 open parenthesis the fraction 1 over 2 close parenthesis squared plus 9 times the fraction 1 over 2 minus 5, equals ?, 0; 2 times the fraction 1 over 4 plus 9 times the fraction 1 over 2 minus 5, equals ?, 0; the fraction 1 over 2 plus the fraction 9 over 2 minus 5, equals ?, 0; the fraction 10 over 2 minus 5, equals ?, 0; 5 minus 5, equals ?, 0; 0, equals, 0 ✓")[$2 x^(2) + 9 x − 5 & = & 0 \ 2 attach(( frac(1, 2) ), t: 2) + 9 ⋅ frac(1, 2) − 5 & limits(=)^(?) & 0 \ 2 ⋅ frac(1, 4) + 9 ⋅ frac(1, 2) − 5 & limits(=)^(?) & 0 \ frac(1, 2) + frac(9, 2) − 5 & limits(=)^(?) & 0 \ frac(10, 2) − 5 & limits(=)^(?) & 0 \ 5 − 5 & limits(=)^(?) & 0 \ 0 & = & 0 ✓$]], [], [Substitute #math.equation(block: false, alt: "x equals minus 5")[$x = − 5$].], [#math.equation(block: false, alt: "2 x squared plus 9 x minus 5, equals, 0; 2 open parenthesis minus 5 close parenthesis squared plus 9 open parenthesis minus 5 close parenthesis minus 5, equals ?, 0; 2 times 25 minus 45 minus 5, equals ?, 0; 50 minus 45 minus 5, equals ?, 0; 0, equals, 0 ✓")[$2 x^(2) + 9 x − 5 & = & 0 \ 2 attach(( − 5 ), t: 2) + 9 ( − 5 ) − 5 & limits(=)^(?) & 0 \ 2 ⋅ 25 − 45 − 5 & limits(=)^(?) & 0 \ 50 − 45 − 5 & limits(=)^(?) & 0 \ 0 & = & 0 ✓$]], )) ] ] #notebox("Video", rgb("#DC2626"), rgb("#DC2626"), rgb("#f7f8fa"))[ #link("https://openstax.org/r/Solving_Quadratics")[Solving Quadratics with the Quadratic Formula] ] === Solving Real-World Applications Modeled by Quadratic Equations There are problem solving strategies that will work well for applications that translate to quadratic equations. Here’s a problem-solving strategy to solve word problems: #strong[Step 1:] Read the problem. Make sure all the words and ideas are understood. #strong[Step 2:] Identify what we are looking for. #strong[Step 3:] Name what we are looking for. Choose a variable to represent that quantity. #strong[Step 4:] Translate into an equation. It may be helpful to restate the problem in one sentence with all the important information. Then, translate the English sentence into an algebra equation. #strong[Step 5:] Solve the equation using good algebra techniques. #strong[Step 6:] Check the answer in the problem and make sure it makes sense. #strong[Step 7:] Answer the question with a complete sentence. #examplebox("Example 13")[Finding Consecutive Integers][ The product of two consecutive integers is 132. Find the integers. #solutionbox[ #strong[Step 1:] Read the problem. #strong[Step 2:] Identify what we are looking for. We are looking for two consecutive integers. #strong[Step 3:] Name what we are looking for. Let #math.equation(block: false, alt: "n equals")[$n =$] the first integer Let #math.equation(block: false, alt: "n plus 1 equals")[$n + 1 =$] the next consecutive integer. #strong[Step 4:] Translate into an equation. Restate the problem in a sentence. #math.equation(block: true, alt: "n open parenthesis n plus 1 close parenthesis equals 132")[$n ( n + 1 ) = 132$] The product of the two consecutive integers is 132. The first integer times the next integer is 132. #strong[Step 5:] Solve the equation. #math.equation(block: true, alt: "n squared plus n equals 132")[$n^(2) + n = 132$] Bring all the terms to one side. #math.equation(block: true, alt: "n squared plus n minus 132 equals 0")[$n^(2) + n − 132 = 0$] Factor the trinomial. #math.equation(block: true, alt: "open parenthesis n minus 11 close parenthesis open parenthesis n plus 12 close parenthesis equals 0")[$( n − 11 ) ( n + 12 ) = 0$] Use the zero product property. #math.equation(block: true, alt: "n minus 11 equals 0 or n plus 12 equals 0")[$n − 11 = 0 #h(0.28em) "or" #h(0.28em) n + 12 = 0$] Solve the equations. #math.equation(block: true, alt: "n equals 11 , n equals minus 12")[$n = 11 , n = − 12$] There are two values for #math.equation(block: false, alt: "n")[$n$] that are solutions to this problem. So, there are two sets of consecutive integers that will work. If the first integer is #math.equation(block: false, alt: "n equals 11")[$n = 11$], then the next integer is 12. If the first integer is #math.equation(block: false, alt: "n equals minus 12")[$n = − 12$], then the next integer is #math.equation(block: false, alt: "minus 11")[$− 11$]. #strong[Step 6:] Check the answer. The consecutive integers are 11, 12 and #math.equation(block: false, alt: "minus 11")[$− 11$], #math.equation(block: false, alt: "minus 12")[$− 12$]. The product of 11 and #math.equation(block: false, alt: "12 equals 132")[$12 = 132$] and the product of #math.equation(block: false, alt: "minus 11 open parenthesis minus 12 close parenthesis equals 132")[$− 11 ( − 12 ) = 132$]. Both pairs of consecutive integers are solutions. #strong[Step 7:] Answer the question. The consecutive integers are 11, 12, and #math.equation(block: false, alt: "minus 11")[$− 11$], #math.equation(block: false, alt: "minus 12")[$− 12$]. ] ] Were you surprised by the pair of negative integers that is one of the solutions? In some applications, negative solutions will result from the algebra, but will not be realistic for the situation. #examplebox("Example 14")[Finding Length and Width of a Garden][ A rectangular garden has an area 15 square feet. The length of the garden is 2 feet more than the width. Find the length and width of the garden. #solutionbox[ #strong[Step 1:] Read the problem. In problems involving geometric figures, a sketch can help you visualize the situation . #figure(figph[A rectangular garden with its length and width marked W plus 2 and W.], alt: "A rectangular garden with its length and width marked W plus 2 and W.", caption: none) #strong[Step 2:] Identify what you are looking for. We are looking for the length and width. #strong[Step 3:] Name what you are looking for. The length is 2 feet more than width. Let #math.equation(block: false, alt: "W equals")[$W =$] the width of the garden. #math.equation(block: false, alt: "W plus 2 equals")[$W + 2 =$] the length of the garden. #strong[Step 4:] Translate into an equation. Restate the important information in a sentence. The area of the rectangular garden is 15 square feet. Use the formula for the area of a rectangle. #math.equation(block: true, alt: "A equals L times W")[$A = L × W$] Substitute in the variables. #math.equation(block: true, alt: "15 equals open parenthesis W plus 2 close parenthesis W")[$15 = ( W + 2 ) W$] #strong[Step 5:] Solve the equation. Distribute first. #math.equation(block: true, alt: "15 equals W squared plus 2 W")[$15 = W^(2) + 2 W$] Get zero on one side. #math.equation(block: true, alt: "0 equals W squared plus 2 W minus 15")[$0 = W^(2) + 2 W − 15$] Factor the trinomial. #math.equation(block: true, alt: "0 equals open parenthesis W plus 5 close parenthesis open parenthesis W minus 3 close parenthesis")[$0 = ( W + 5 ) ( W − 3 )$] Use the Zero Product Property. #math.equation(block: true, alt: "0 equals W plus 5")[$0 = W + 5$] #math.equation(block: true, alt: "0 equals W minus 3")[$0 = W − 3$] Solve each equation. #math.equation(block: true, alt: "minus 5 equals W")[$− 5 = W$] #math.equation(block: true, alt: "3 equals W")[$3 = W$] Since #math.equation(block: false, alt: "W")[$W$] is the width of the garden, it does not make sense for it to be negative. We eliminate that value for #math.equation(block: false, alt: "W")[$W$]. #math.equation(block: true, alt: "W equals 3")[$W = 3$] Width is 3 feet. Find the value of the length. #math.equation(block: false, alt: "W plus 2 equals")[$W + 2 =$] length. #math.equation(block: true, alt: "3 plus 2")[$3 + 2$] 5 Length is 5 feet. #strong[Step 6:] Check the answer . Does the answer make sense? #figure(figph[A rectangular garden with its length and width marked W plus 2 and W. The length is marked 3 plus 2 equals 5. The width is marked 3. Text reads, A equals L times W. A equals 3 times 5. A equals 15.], alt: "A rectangular garden with its length and width marked W plus 2 and W. The length is marked 3 plus 2 equals 5. The width is marked 3. Text reads, A equals L times W. A equals 3 times 5. A equals 15.", caption: none) Yes, this makes sense. #strong[Step 7:] Answer the question. The width of the garden is 3 feet and the length is 5 feet. ] ] #notebox("Work It Out", rgb("#059669"), rgb("#007942"), rgb("#EAF3EC"))[ #emph[Completing the Square] Recall the two methods used to solve quadratic equations of the form #math.equation(block: false, alt: "a x squared plus b x plus c :")[$a x^(2) + b x + c :$] by factoring and by using the quadratic formula. There are, however, many different methods for solving quadratic equations that were developed throughout history. Egyptian, Mesopotamian, Chinese, Indian, and Greek mathematicians all solved various types of quadratic equations, as did Arab mathematicians of the ninth through the twelfth centuries. It is one of these Arab mathematicians' methods that we wish to investigate with this activity. Muhammad ibn Musa al-Khwarizmi was employed as a scholar at the House of Wisdom in Baghdad, located in present day Iraq. One of the many accomplishments of Al-Khwarizmi was his book on the topic of algebra. In that book, he asks, “What must be the square which, when increased by ten of its own roots, amounts to 39?” Al-Khwarizmi, like many Arab mathematicians of his time, was well versed in Euclid's Elements. Like Euclid, he viewed algebra very geometrically, and thus had a geometric approach to solving a problem like the one above. In his approach, he used a method which today we refer to as #emph[completing the square]. His description of the solution method for the above problem is: halve the number of roots, which in the present instance yields 5. This you multiply by itself; the product is 25. Add this to 39; the sum is 64. Now take the root of this which is 8, and subtract from it half the number of the roots, which is 5; the remainder is 3. This is the root of the square which you sought for. Thus the square is 9. #figure(table( columns: 2, align: left, inset: 6pt, table.header([So, what does all of this mean? Al-Khwarizmi would start with a square of unknown length of side (we will label the side length #math.equation(block: false, alt: "x")[$x$]). See So, this square has area #math.equation(block: false, alt: "x squared")[$x^(2)$]], [#figure(figph[A light blue square with a thick black border sits above a black lowercase 'x' on a white background.], alt: "A light blue square with a thick black border sits above a black lowercase 'x' on a white background.", caption: none) #math.equation(block: false, alt: "x")[$x$]]), [He would then proceed to halve the number of roots (i.e., there are 10 roots by which the square is increasing) to get 5; this he would add to the first square. See The area of the two new pieces added into the original square are both #math.equation(block: false, alt: "5 x")[$5 x$]. At this point, we have #math.equation(block: false, alt: "x squared plus 10 x equals 39")[$x^(2) + 10 x = 39$].], [#figure( stdfig([#cetz.canvas({ import cetz.draw: * line((0.00pt, 0.00pt), (80.00pt, 0.00pt), (80.00pt, 80.00pt), (0.00pt, 80.00pt), close: true, fill: none, stroke: 0.7pt + black) content((40.00pt, -9.00pt), anchor: "north", text(size: 9pt)[x]) content((-9.00pt, 40.00pt), anchor: "east", text(size: 9pt)[x]) line((80.00pt, 0.00pt), (180.00pt, 0.00pt), (180.00pt, 80.00pt), (80.00pt, 80.00pt), close: true, fill: rgb("#e5e7eb"), stroke: 0.7pt + black) content((130.00pt, -9.00pt), anchor: "north", text(size: 9pt)[5]) content((189.00pt, 40.00pt), anchor: "west", text(size: 9pt)[x]) content((130.00pt, 89.00pt), anchor: "south", text(size: 9pt)[5]) })]), alt: "A square with a shaded rectangle attached to its right and bottom sides. The sides of the square measure x. The length and width of the rectangle measure x and 5.", caption: none, )], [Now Al-Khwarizmi needed to “complete the square” by adding into the drawing a small square. See This square has an area of 25.], [#figure( stdfig([#cetz.canvas({ import cetz.draw: * line((16.67pt, 100.00pt), (100.00pt, 100.00pt), (100.00pt, 183.33pt), (16.67pt, 183.33pt), close: true, fill: none, stroke: 0.7pt + black) content((58.33pt, 91.00pt), anchor: "north", text(size: 9pt)[x]) content((109.00pt, 141.67pt), anchor: "west", text(size: 9pt)[x]) content((58.33pt, 192.33pt), anchor: "south", text(size: 9pt)[x]) content((7.67pt, 141.67pt), anchor: "east", text(size: 9pt)[x]) line((100.00pt, 100.00pt), (183.33pt, 100.00pt), (183.33pt, 183.33pt), (100.00pt, 183.33pt), close: true, fill: none, stroke: 0.7pt + black) content((141.67pt, 91.00pt), anchor: "north", text(size: 9pt)[5]) content((192.33pt, 141.67pt), anchor: "west", text(size: 9pt)[x]) content((141.67pt, 192.33pt), anchor: "south", text(size: 9pt)[5]) line((16.67pt, 16.67pt), (100.00pt, 16.67pt), (100.00pt, 100.00pt), (16.67pt, 100.00pt), close: true, fill: rgb("#e5e7eb"), stroke: 0.7pt + black) content((58.33pt, 7.67pt), anchor: "north", text(size: 9pt)[x]) content((109.00pt, 58.33pt), anchor: "west", text(size: 9pt)[5]) content((7.67pt, 58.33pt), anchor: "east", text(size: 9pt)[5]) line((100.00pt, 16.67pt), (183.33pt, 16.67pt), (183.33pt, 100.00pt), (100.00pt, 100.00pt), close: true, fill: none, stroke: (paint: black, thickness: 0.6pt, dash: "dashed")) })]), alt: "A square with a rectangle attached to its right and bottom sides. The sides of the square measure x. The length and width of the rectangle measure x and 5. The rectangle at the bottom is shaded. A square in dashed lines with rounded edges is present at the bottom-right corner.", caption: none, )], )) #math.equation(block: false, alt: "x squared plus 10 x plus 25 equals 39 plus 25")[$x^(2) + 10 x + 25 = 39 + 25$], or #math.equation(block: false, alt: "x squared plus 10 x plus 25 equals 64")[$x^(2) + 10 x + 25 = 64$]. Notice that the completed square has side length #math.equation(block: false, alt: "x plus 5")[$x + 5$], so the large square has area #math.equation(block: false, alt: "open parenthesis x plus 5 close parenthesis squared")[$attach(( x + 5 ), t: 2)$]. (Notice algebraically that the left half of the equation #math.equation(block: false, alt: "x squared plus 10 x plus 25 equals 64")[$x^(2) + 10 x + 25 = 64$] factors to #math.equation(block: false, alt: "open parenthesis x plus 5 close parenthesis squared equals 64. close parenthesis")[$attach(\( x + 5 \), t: 2) = 64. \)$] This means the area of large square equals 64. If #math.equation(block: false, alt: "open parenthesis x plus 5 close parenthesis squared equals 64")[$attach(( x + 5 ), t: 2) = 64$], then #math.equation(block: false, alt: "x plus 5 equals 8")[$x + 5 = 8$]; so #math.equation(block: false, alt: "x")[$x$] must be equal to 3 or #math.equation(block: false, alt: "minus 13")[$− 13$] to make this true. Note that Al-Kwarimi would not have considered the possibility of a negative solution, since he approached the solution geometrically, and negative distances do not exist. ] === Key Terms - monomial - polynomial - binomial - trinomial - quadratic equation - Zero Product Property === Key Concepts - A quadratic equation is an algebraic equation where the highest power (degree) of the equation is two. - To solve a quadratic equation is to find the value(s) that when substituted in for the variables, will make the equation equal to zero. - There can be two, one, or no solutions to any quadratic equation. - There are several methods to solve a quadratic equation. These methods include factoring quadratic equations, graphic quadratic equations, using the square root method, and using the quadratic formula. === Videos - #link("https://openstax.org/r/Factoring_with_the_Box")[Factoring with the Box Method (Area Model)] - #link("https://openstax.org/r/Zero_Property")[Solving Quadratics with the Zero Property] - #link("https://openstax.org/r/Solving_Quadratics")[Solving Quadratics with the Quadratic Formula]