#set document(title: "9.4 Solve Equations in Quadratic Form", 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"))


== 9.4#h(0.6em)Solve Equations in Quadratic Form

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Before you get started, take this readiness quiz.

Factor by substitution: #math.equation(block: false, alt: "y to the power 4 minus y squared minus 20 .")[$y^(4) − y^(2) − 20 .$] #linebreak() 
If you missed this problem, review .

#solutionbox[
#math.equation(block: true, alt: "y squared plus 4 y squared minus 5")[$y^(2) + 4 y^(2) − 5$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Factor by substitution: #math.equation(block: false, alt: "open parenthesis y minus 4 close parenthesis squared plus 8 open parenthesis y minus 4 close parenthesis plus 15 .")[$attach(( y − 4 ), t: 2) + 8 ( y − 4 ) + 15 .$] #linebreak() 
If you missed this problem, review .

#solutionbox[
#math.equation(block: true, alt: "open parenthesis y minus 1 close parenthesis open parenthesis y plus 1 close parenthesis")[$( y − 1 ) ( y + 1 )$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Simplify: ⓐ #math.equation(block: false, alt: "x to the power the fraction 1 over 2 times x to the power the fraction 1 over 4")[$x^(frac(1, 2)) · x^(frac(1, 4))$] ⓑ #math.equation(block: false, alt: "open parenthesis x to the power the fraction 1 over 3 close parenthesis squared")[$attach(( x^(frac(1, 3)) ), t: 2)$] ⓒ #math.equation(block: false, alt: "open parenthesis x to the power −1 close parenthesis squared .")[$attach(( x^(−1) ), t: 2) .$] #linebreak() 
If you missed this problem, review .

#solutionbox[
ⓐ #math.equation(block: false, alt: "x to the power the fraction 3 over 4")[$x^(frac(3, 4))$]; ⓑ #math.equation(block: false, alt: "x to the power the fraction 2 over 3")[$x^(frac(2, 3))$]; ⓒ #math.equation(block: false, alt: "x to the power minus 2")[$x^(− 2)$]
]
]


=== Solve Equations in Quadratic Form

Sometimes when we factored trinomials, the trinomial did not appear to be in the #emph[ax]#super[2] + #emph[bx] + #emph[c] form. So we factored by substitution allowing us to make it fit the #emph[ax]#super[2] + #emph[bx] + #emph[c] form. We used the standard #math.equation(block: false, alt: "u")[$u$] for the substitution.

To factor the expression #emph[x]#super[4] − 4#emph[x]#super[2] − 5, we noticed the variable part of the middle term is #emph[x]#super[2] and its square, #emph[x]#super[4], is the variable part of the first term. (We know #math.equation(block: false, alt: "open parenthesis x squared close parenthesis squared equals x to the power 4 .")[$attach(( x^(2) ), t: 2) = x^(4) .$]) So we let #emph[u] = #emph[x]#super[2] and factored.

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#figure(figph[The image displays the mathematical expression x^4 - 4x^2 - 5, a polynomial equation often seen in algebra and calculus.], alt: "The image displays the mathematical expression x^4 - 4x^2 - 5, a polynomial equation often seen in algebra and calculus.", caption: none)]),
  [], [#figure(figph[The mathematical expression (x^2)^2 - 4(x^2) - 5 is shown, representing a quadratic form in terms of x^2. The 'x' variable and its exponents are highlighted in red.], alt: "The mathematical expression (x^2)^2 - 4(x^2) - 5 is shown, representing a quadratic form in terms of x^2. The 'x' variable and its exponents are highlighted in red.", caption: none)],
  [Let #math.equation(block: false, alt: "u equals x squared")[$u = x^(2)$] and substitute.], [#figure(figph[The image shows the mathematical expression u^2 - 4u - 5, where 'u' is highlighted in red, indicating it as a variable.], alt: "The image shows the mathematical expression u^2 - 4u - 5, where 'u' is highlighted in red, indicating it as a variable.", caption: none)],
  [Factor the trinomial.], [#figure(figph[The mathematical expression (u + 1)(u - 5) is displayed on a white background.], alt: "The mathematical expression (u + 1)(u - 5) is displayed on a white background.", caption: none)],
  [Replace #emph[u] with #math.equation(block: false, alt: "x squared")[$x^(2)$].], [#figure(figph[A mathematical expression showing the product of two binomials: (x^2 + 1)(x^2 - 5). The variable 'x' is in red, while the numbers, operators, and parentheses are in black.], alt: "A mathematical expression showing the product of two binomials: (x^2 + 1)(x^2 - 5). The variable 'x' is in red, while the numbers, operators, and parentheses are in black.", caption: none)],
))

Similarly, sometimes an equation is not in the #emph[ax]#super[2] + #emph[bx] + #emph[c] = 0 form but looks much like a quadratic equation. Then, we can often make a thoughtful substitution that will allow us to make it fit the #emph[ax]#super[2] + #emph[bx] + #emph[c] = 0 form. If we can make it fit the form, we can then use all of our methods to solve quadratic equations.

Notice that in the quadratic equation #emph[ax]#super[2] + #emph[bx] + #emph[c] = 0, the middle term has a variable, #emph[x], and its square, #emph[x]#super[2], is the variable part of the first term. Look for this relationship as you try to find a substitution.

Again, we will use the standard #emph[u] to make a substitution that will put the equation in quadratic form. If the substitution gives us an equation of the form #emph[ax]#super[2] + #emph[bx] + #emph[c] = 0, we say the original equation was of #strong[quadratic form].

The next example shows the steps for solving an equation in quadratic form.

#examplebox("Example 1")[How to Solve Equations in Quadratic Form][
Solve: #math.equation(block: false, alt: "6 x to the power 4 minus 7 x squared plus 2 equals 0")[$6 x^(4) − 7 x^(2) + 2 = 0$]

#solutionbox[
#figure(figph[Step 1 is to identify a substitution that will put the equation in quadratic form. Look at the equation 6 x to the fourth power minus 7 x squared plus 2 equals 0. Since the square of x squared equals x to the fourth, let u equals x squared.], alt: "Step 1 is to identify a substitution that will put the equation in quadratic form. Look at the equation 6 x to the fourth power minus 7 x squared plus 2 equals 0. Since the square of x squared equals x to the fourth, let u equals x squared.", caption: none)

#figure(figph[Step 2 is to rewrite the equation with the substitution to put it in quadratic form. Rewrite the equation to prepare for the substitution to show 6 times the square of x squared minus 7 times x squared plus 2 equals 0. Substitute u equals x squared to get the new equation 6 times u squared minus 7 u plus 2 equals 0.], alt: "Step 2 is to rewrite the equation with the substitution to put it in quadratic form. Rewrite the equation to prepare for the substitution to show 6 times the square of x squared minus 7 times x squared plus 2 equals 0. Substitute u equals x squared to get the new equation 6 times u squared minus 7 u plus 2 equals 0.", caption: none)

#figure(figph[Step 3 is to solve the quadratic equation for u. We can solve by factoring, so rewrite the equation as the product of 2 u minus 1 and 3 u minus 2 equals 0. Use the Zero Product Property to create 2 equations. If 2 u minus 1 equals 0, then 2u equals one, so u equals one half. If 3 u minus 2 equals 0, then 3 u equals 2 and u equals two thirds.], alt: "Step 3 is to solve the quadratic equation for u. We can solve by factoring, so rewrite the equation as the product of 2 u minus 1 and 3 u minus 2 equals 0. Use the Zero Product Property to create 2 equations. If 2 u minus 1 equals 0, then 2u equals one, so u equals one half. If 3 u minus 2 equals 0, then 3 u equals 2 and u equals two thirds.", caption: none)

#figure(figph[In step 4, substitute the original variable back into the results. In this case replace u with x squared. So u equals one half becomes x squared equals one half and u equals two thirds becomes x squared equals two thirds.], alt: "In step 4, substitute the original variable back into the results. In this case replace u with x squared. So u equals one half becomes x squared equals one half and u equals two thirds becomes x squared equals two thirds.", caption: none)

#figure(figph[Step 5 is to solve for the original variable, so use the Square Root Property to solve for x. If x squared equals one half, then x equals the positive or negative square root of one half. Rationalize the denominator to see that x equals the positive or negative square root of 2 divided by 2. If x squared equals two thirds, then x equals the positive or negative square root of two thirds. Rationalize the denominator to see that x equals the positive or negative square root of 6 divided by 3.], alt: "Step 5 is to solve for the original variable, so use the Square Root Property to solve for x. If x squared equals one half, then x equals the positive or negative square root of one half. Rationalize the denominator to see that x equals the positive or negative square root of 2 divided by 2. If x squared equals two thirds, then x equals the positive or negative square root of two thirds. Rationalize the denominator to see that x equals the positive or negative square root of 6 divided by 3.", caption: none)

#figure(figph[In step 6, check your solutions. We will show one check here, x equals square root 2 divided by 2. Substitute this value into the original equation. 6 times the fourth power of the quotient square root 2 divided by 2 minus 7 times the square of the quotient square root of 2 divided by 2 plus 2. Does this expression equal 0? Simplify the powers. 6 times four sixteenths minus 7 times two fourths plus 2. Simplify terms. Three halves minus seven halves plus four halves equals zero. Square root 2 divided by 2 is a solution. We leave the other checks to you!], alt: "In step 6, check your solutions. We will show one check here, x equals square root 2 divided by 2. Substitute this value into the original equation. 6 times the fourth power of the quotient square root 2 divided by 2 minus 7 times the square of the quotient square root of 2 divided by 2 plus 2. Does this expression equal 0? Simplify the powers. 6 times four sixteenths minus 7 times two fourths plus 2. Simplify terms. Three halves minus seven halves plus four halves equals zero. Square root 2 divided by 2 is a solution. We leave the other checks to you!", caption: none)
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve: #math.equation(block: false, alt: "x to the power 4 minus 6 x squared plus 8 equals 0")[$x^(4) − 6 x^(2) + 8 = 0$].

#solutionbox[
#math.equation(block: true, alt: "x equals the square root of 2 , x equals − the square root of 2 , x equals 2 , x equals −2")[$x = sqrt(2) , x = "−" sqrt(2) , x = 2 , x = −2$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve: #math.equation(block: false, alt: "x to the power 4 minus 11 x squared plus 28 equals 0")[$x^(4) − 11 x^(2) + 28 = 0$].

#solutionbox[
#math.equation(block: true, alt: "x equals the square root of 7 , x equals − the square root of 7 , x equals 2 , x equals −2")[$x = sqrt(7) , x = "−" sqrt(7) , x = 2 , x = −2$]
]
]

We summarize the steps to solve an equation in quadratic form.

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Solve equations in quadratic form.]

+ Identify a substitution that will put the equation in quadratic form.
+ Rewrite the equation with the substitution to put it in quadratic form.
+ Solve the quadratic equation for #emph[u].
+ Substitute the original variable back into the results, using the substitution.
+ Solve for the original variable.
+ Check the solutions.
]

In the next example, the binomial in the middle term, (#emph[x] − 2) is squared in the first term. If we let #emph[u] = #emph[x] − 2 and substitute, our trinomial will be in #emph[ax]#super[2] + #emph[bx] + #emph[c] form.

#examplebox("Example 2")[][
Solve: #math.equation(block: false, alt: "open parenthesis x minus 2 close parenthesis squared plus 7 open parenthesis x minus 2 close parenthesis plus 12 equals 0 .")[$attach(( x − 2 ), t: 2) + 7 ( x − 2 ) + 12 = 0 .$]

#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#figure(figph[A quadratic equation (x-2)^2 + 7(x-2) + 12 = 0 is displayed on a white background, representing a math problem typically solved using substitution.], alt: "A quadratic equation (x-2)^2 + 7(x-2) + 12 = 0 is displayed on a white background, representing a math problem typically solved using substitution.", caption: none)]),
  [Prepare for the substitution.], [#figure(figph[Equation showing the expression (x minus 2) squared plus 7 times (x minus 2) plus 12 equals zero.], alt: "Equation showing the expression (x minus 2) squared plus 7 times (x minus 2) plus 12 equals zero.", caption: none)],
  [Let #math.equation(block: false, alt: "u equals x minus 2")[$u = x − 2$] and substitute.], [#figure(figph[A quadratic equation displayed on a white background: u^2 + 7u + 12 = 0. The variable 'u' appears in red, while the numbers, exponents, and operators are in black.], alt: "A quadratic equation displayed on a white background: u^2 + 7u + 12 = 0. The variable 'u' appears in red, while the numbers, exponents, and operators are in black.", caption: none)],
  [Solve by factoring.], [#figure(figph[A mathematical equation is displayed against a white background. The equation reads as follows: (u + 3)(u + 4) = 0.], alt: "A mathematical equation is displayed against a white background. The equation reads as follows: (u + 3)(u + 4) = 0.", caption: none)

 #linebreak() #figure(figph[Two mathematical equations are displayed: u + 3 = 0, u + 4 = 0, shown in a white background with black text.], alt: "Two mathematical equations are displayed: u + 3 = 0, u + 4 = 0, shown in a white background with black text.", caption: none)

 #linebreak() #figure(figph[The image displays two mathematical expressions: u = -3, and u = -4, written in a clear, sans-serif font against a plain white background.], alt: "The image displays two mathematical expressions: u = -3, and u = -4, written in a clear, sans-serif font against a plain white background.", caption: none)],
  [Replace #math.equation(block: false, alt: "u")[$u$] with #math.equation(block: false, alt: "x minus 2 .")[$x − 2 .$]], [#figure(figph[Two mathematical equations are displayed: x - 2 = -3 and x - 2 = -4. These are linear equations where a variable 'x' is involved in a subtraction operation resulting in a negative integer.], alt: "Two mathematical equations are displayed: x - 2 = -3 and x - 2 = -4. These are linear equations where a variable 'x' is involved in a subtraction operation resulting in a negative integer.", caption: none)],
  [Solve for #math.equation(block: false, alt: "x .")[$x .$]], [#figure(figph[Two mathematical equations are displayed on a white background: 'X = -1' followed by a comma, and then 'X = -2'.], alt: "Two mathematical equations are displayed on a white background: 'X = -1' followed by a comma, and then 'X = -2'.", caption: none)],
  [Check: #linebreak()  #linebreak() 
#figure(figph[Verification of solutions for a quadratic equation (x-2)^2 + 7(x-2) + 12 = 0, showing that both x=-1 and x=-2 satisfy the equation.], alt: "Verification of solutions for a quadratic equation (x-2)^2 + 7(x-2) + 12 = 0, showing that both x=-1 and x=-2 satisfy the equation.", caption: none)], [],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve: #math.equation(block: false, alt: "open parenthesis x minus 5 close parenthesis squared plus 6 open parenthesis x minus 5 close parenthesis plus 8 equals 0 .")[$attach(( x − 5 ), t: 2) + 6 ( x − 5 ) + 8 = 0 .$]

#solutionbox[
#math.equation(block: true, alt: "x equals 3 , x equals 1")[$x = 3 , x = 1$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve: #math.equation(block: false, alt: "open parenthesis y minus 4 close parenthesis squared plus 8 open parenthesis y minus 4 close parenthesis plus 15 equals 0 .")[$attach(( y − 4 ), t: 2) + 8 ( y − 4 ) + 15 = 0 .$]

#solutionbox[
#math.equation(block: true, alt: "y equals −1 , y equals 1")[$y = −1 , y = 1$]
]
]

In the next example, we notice that #math.equation(block: false, alt: "open parenthesis the square root of x close parenthesis squared equals x .")[$attach(( sqrt(x) ), t: 2) = x .$] Also, remember that when we square both sides of an equation, we may introduce extraneous roots. Be sure to check your answers!

#examplebox("Example 3")[][
Solve: #math.equation(block: false, alt: "x minus 3 the square root of x plus 2 equals 0 .")[$x − 3 sqrt(x) + 2 = 0 .$]

#solutionbox[
The #math.equation(block: false, alt: "the square root of x")[$sqrt(x)$] in the middle term, is squared in the first term #math.equation(block: false, alt: "open parenthesis the square root of x close parenthesis squared equals x .")[$attach(( sqrt(x) ), t: 2) = x .$] If we let #math.equation(block: false, alt: "u equals the square root of x")[$u = sqrt(x)$] and substitute, our trinomial will be in #emph[ax]#super[2] + #emph[bx] + #emph[c] = 0 form.

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#figure(figph[A mathematical equation is displayed: x - 3 times the square root of x + 2 = 0. This is a radical equation that can be solved by substitution, letting y = square root of x.], alt: "A mathematical equation is displayed: x - 3 times the square root of x + 2 = 0. This is a radical equation that can be solved by substitution, letting y = square root of x.", caption: none)]),
  [Rewrite the trinomial to prepare for the substitution.], [#figure(figph[A quadratic equation, (sqrt(x))^2 - 3sqrt(x) + 2 = 0, is shown, with the square root of x highlighted in red.], alt: "A quadratic equation, (sqrt(x))^2 - 3sqrt(x) + 2 = 0, is shown, with the square root of x highlighted in red.", caption: none)],
  [Let #math.equation(block: false, alt: "u equals the square root of x")[$u = sqrt(x)$] and substitute.], [#figure(figph[A white background displays the quadratic equation u^2 - 3u + 2 = 0 in horizontal alignment. The variable 'u' is rendered in red, while the numbers, operators, and equals sign are black.], alt: "A white background displays the quadratic equation u^2 - 3u + 2 = 0 in horizontal alignment. The variable 'u' is rendered in red, while the numbers, operators, and equals sign are black.", caption: none)],
  [Solve by factoring.], [#figure(figph[A mathematical equation is displayed on a white background: (u-2)(u-1)=0.], alt: "A mathematical equation is displayed on a white background: (u-2)(u-1)=0.", caption: none)

 #linebreak() #figure(figph[Two mathematical equations are displayed: u - 2 = 0, u - 1 = 0, separated by a comma. The equations are presented in a clear, sans-serif font against a plain white background.], alt: "Two mathematical equations are displayed: u - 2 = 0, u - 1 = 0, separated by a comma. The equations are presented in a clear, sans-serif font against a plain white background.", caption: none)

 #linebreak() #figure(figph[The image displays mathematical notation 'u = 2, u = 1' in black text against a white background.], alt: "The image displays mathematical notation 'u = 2, u = 1' in black text against a white background.", caption: none)],
  [Replace #emph[u] with #math.equation(block: false, alt: "the square root of x .")[$sqrt(x) .$]], [#figure(figph[Two mathematical equations are displayed: square root of x equals 2, and square root of x equals 1.], alt: "Two mathematical equations are displayed: square root of x equals 2, and square root of x equals 1.", caption: none)],
  [Solve for #emph[x], by squaring both sides.], [#figure(figph[The text 'x = 4, x = 1' is shown against a white background.], alt: "The text 'x = 4, x = 1' is shown against a white background.", caption: none)],
  [Check: #linebreak()  #linebreak() 
#figure(figph[Checking the solutions x=4 and x=1 for the equation x - 3sqrt(x) + 2 = 0, demonstrating that both values correctly satisfy the equality.], alt: "Checking the solutions x=4 and x=1 for the equation x - 3sqrt(x) + 2 = 0, demonstrating that both values correctly satisfy the equality.", caption: none)], [],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve: #math.equation(block: false, alt: "x minus 7 the square root of x plus 12 equals 0 .")[$x − 7 sqrt(x) + 12 = 0 .$]

#solutionbox[
#math.equation(block: true, alt: "x equals 9 , x equals 16")[$x = 9 , x = 16$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve: #math.equation(block: false, alt: "x minus 6 the square root of x plus 8 equals 0 .")[$x − 6 sqrt(x) + 8 = 0 .$]

#solutionbox[
#math.equation(block: true, alt: "x equals 4 , x equals 16")[$x = 4 , x = 16$]
]
]

Substitutions for rational exponents can also help us solve an equation in quadratic form. Think of the properties of exponents as you begin the next example.

#examplebox("Example 4")[][
Solve: #math.equation(block: false, alt: "x to the power the fraction 2 over 3 minus 2 x to the power the fraction 1 over 3 minus 24 equals 0 .")[$x^(frac(2, 3)) − 2 x^(frac(1, 3)) − 24 = 0 .$]

#solutionbox[
The #math.equation(block: false, alt: "x to the power the fraction 1 over 3")[$x^(frac(1, 3))$] in the middle term is squared in the first term #math.equation(block: false, alt: "open parenthesis x to the power the fraction 1 over 3 close parenthesis squared equals x to the power the fraction 2 over 3 .")[$attach(( x^(frac(1, 3)) ), t: 2) = x^(frac(2, 3)) .$] If we let #math.equation(block: false, alt: "u equals x to the power the fraction 1 over 3")[$u = x^(frac(1, 3))$] and substitute, our trinomial will be in #emph[ax]#super[2] + #emph[bx] + #emph[c] = 0 form.

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#figure(figph[A mathematical equation is displayed, reading 'x to the power of 2/3 minus 2x to the power of 1/3 minus 24 equals 0'.], alt: "A mathematical equation is displayed, reading 'x to the power of 2/3 minus 2x to the power of 1/3 minus 24 equals 0'.", caption: none)]),
  [Rewrite the trinomial to prepare for the substitution.], [#figure(figph[A mathematical equation is displayed: (x raised to the power of 1/3) squared minus 2 times (x raised to the power of 1/3) minus 24 equals 0.], alt: "A mathematical equation is displayed: (x raised to the power of 1/3) squared minus 2 times (x raised to the power of 1/3) minus 24 equals 0.", caption: none)],
  [Let #math.equation(block: false, alt: "u equals x to the power the fraction 1 over 3")[$u = x^(frac(1, 3))$] and substitute.], [#figure(figph[A quadratic equation is displayed, showing u^2 - 2u - 24 = 0. The variable 'u' and its coefficient '2' in the second term are highlighted in red.], alt: "A quadratic equation is displayed, showing u^2 - 2u - 24 = 0. The variable 'u' and its coefficient '2' in the second term are highlighted in red.", caption: none)],
  [Solve by factoring.], [#figure(figph[A mathematical equation in factored form is shown against a white background: (u - 6)(u + 4) = 0.], alt: "A mathematical equation in factored form is shown against a white background: (u - 6)(u + 4) = 0.", caption: none)

 #linebreak() #figure(figph[Two mathematical equations are displayed: u - 6 = 0 and u + 4 = 0. These linear equations show simple algebraic expressions involving the variable 'u' and constants, both set to zero.], alt: "Two mathematical equations are displayed: u - 6 = 0 and u + 4 = 0. These linear equations show simple algebraic expressions involving the variable 'u' and constants, both set to zero.", caption: none)

 #linebreak() #figure(figph[Two mathematical expressions are shown: u = 6, and u = -4.], alt: "Two mathematical expressions are shown: u = 6, and u = -4.", caption: none)],
  [Replace #emph[u] with #math.equation(block: false, alt: "x to the power the fraction 1 over 3 .")[$x^(frac(1, 3)) .$]], [#figure(figph[Two mathematical equations are shown: x^(1/3) = 6, and x^(1/3) = -4. These represent two separate conditions for x raised to the power of one-third.], alt: "Two mathematical equations are shown: x^(1/3) = 6, and x^(1/3) = -4. These represent two separate conditions for x raised to the power of one-third.", caption: none)],
  [Solve for #math.equation(block: false, alt: "x")[$x$] by cubing both sides.], [#figure(figph[Two algebraic equations are displayed: (x^(1/3))^3 = (6)^3 and (x^(1/3))^3 = (-4)^3, illustrating how to solve for x when involving cube roots and cubing.], alt: "Two algebraic equations are displayed: (x^(1/3))^3 = (6)^3 and (x^(1/3))^3 = (-4)^3, illustrating how to solve for x when involving cube roots and cubing.", caption: none)

 #linebreak() #figure(figph[Two mathematical expressions are displayed on a white background: 'x = 216,' and 'x = -64'.], alt: "Two mathematical expressions are displayed on a white background: 'x = 216,' and 'x = -64'.", caption: none)],
  [Check: #linebreak()  #linebreak() 
#figure(figph[This image demonstrates verifying the solutions x=216 and x=-64 for the equation x^(2/3) - 2x^(1/3) - 24 = 0 by substituting them into the equation, both resulting in 0=0.], alt: "This image demonstrates verifying the solutions x=216 and x=-64 for the equation x^(2/3) - 2x^(1/3) - 24 = 0 by substituting them into the equation, both resulting in 0=0.", caption: none)], [],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve: #math.equation(block: false, alt: "x to the power the fraction 2 over 3 minus 5 x to the power the fraction 1 over 3 minus 14 equals 0 .")[$x^(frac(2, 3)) − 5 x^(frac(1, 3)) − 14 = 0 .$]

#solutionbox[
#math.equation(block: true, alt: "x equals −8 , x equals 343")[$x = −8 , x = 343$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve: #math.equation(block: false, alt: "x to the power the fraction 1 over 2 minus 8 x to the power the fraction 1 over 4 plus 15 equals 0 .")[$x^(frac(1, 2)) − 8 x^(frac(1, 4)) + 15 = 0 .$]

#solutionbox[
#math.equation(block: true, alt: "x equals 81 , x equals 625")[$x = 81 , x = 625$]
]
]

In the next example, we need to keep in mind the definition of a negative exponent as well as the properties of exponents.

#examplebox("Example 5")[][
Solve: #math.equation(block: false, alt: "3 x to the power −2 minus 7 x to the power −1 plus 2 equals 0 .")[$3 x^(−2) − 7 x^(−1) + 2 = 0 .$]

#solutionbox[
The #math.equation(block: false, alt: "x to the power −1")[$x^(−1)$] in the middle term is squared in the first term #math.equation(block: false, alt: "open parenthesis x to the power −1 close parenthesis squared equals x to the power −2 .")[$attach(( x^(−1) ), t: 2) = x^(−2) .$] If we let #math.equation(block: false, alt: "u equals x to the power −1")[$u = x^(−1)$] and substitute, our trinomial will be in #emph[ax]#super[2] + #emph[bx] + #emph[c] = 0 form.

#figure(table(
  columns: 3, align: left, inset: 6pt,
  table.header([], [], [#figure(figph[A mathematical equation displays '3x^(-2) - 7x^(-1) + 2 = 0' in a simple, clear font against a white background.], alt: "A mathematical equation displays '3x^(-2) - 7x^(-1) + 2 = 0' in a simple, clear font against a white background.", caption: none)]),
  [Rewrite the trinomial to prepare for the substitution.], [], [#figure(figph[A mathematical equation is displayed: 3(x⁻¹)² - 7(x⁻¹) + 2 = 0. The variable 'x' is shown in red, with an exponent of -1. The equation features a quadratic form with respect to x⁻¹.], alt: "A mathematical equation is displayed: 3(x⁻¹)² - 7(x⁻¹) + 2 = 0. The variable 'x' is shown in red, with an exponent of -1. The equation features a quadratic form with respect to x⁻¹.", caption: none)],
  [Let #math.equation(block: false, alt: "u equals x to the power −1")[$u = x^(−1)$] and substitute.], [], [#figure(figph[A quadratic equation is displayed on a white background: 3u^2 - 7u + 2 = 0. The coefficients for the u-squared and u terms, '3u' and '7u', are highlighted in red, contrasting with the black numbers and operators.], alt: "A quadratic equation is displayed on a white background: 3u^2 - 7u + 2 = 0. The coefficients for the u-squared and u terms, '3u' and '7u', are highlighted in red, contrasting with the black numbers and operators.", caption: none)],
  [Solve by factoring.], [], [#figure(figph[A mathematical equation where the product of two binomials, (3u - 1) and (u - 2), is set equal to zero.], alt: "A mathematical equation where the product of two binomials, (3u - 1) and (u - 2), is set equal to zero.", caption: none)],
  [], [], [#figure(figph[Two separate linear equations, 3u - 1 = 0 and u - 2 = 0, are shown in white text against a dark background, representing a mathematical problem to solve for the variable 'u'.], alt: "Two separate linear equations, 3u - 1 = 0 and u - 2 = 0, are shown in white text against a dark background, representing a mathematical problem to solve for the variable 'u'.", caption: none)],
  [], [], [#figure(figph[Two mathematical expressions are displayed: u = 1/3, and u = 2.], alt: "Two mathematical expressions are displayed: u = 1/3, and u = 2.", caption: none)],
  [Replace #emph[u] with #math.equation(block: false, alt: "x to the power −1 .")[$x^(−1) .$]], [], [#figure(figph[Two mathematical equations are displayed: x^-1 = 1/3, and x^-1 = 2. Both equations show an inverse power of x set equal to a constant value, with the first constant being a fraction and the second an integer.], alt: "Two mathematical equations are displayed: x^-1 = 1/3, and x^-1 = 2. Both equations show an inverse power of x set equal to a constant value, with the first constant being a fraction and the second an integer.", caption: none)],
  [Solve for #math.equation(block: false, alt: "x")[$x$] by taking the reciprocal since #math.equation(block: false, alt: "x to the power minus 1 equals the fraction 1 over x .")[$x^(− 1) = frac(1, x) .$]], [], [#figure(figph[Two mathematical expressions are displayed: 'x = 3,' and 'x = 1/2'. These represent potential solutions or values for the variable x.], alt: "Two mathematical expressions are displayed: 'x = 3,' and 'x = 1/2'. These represent potential solutions or values for the variable x.", caption: none)],
  [Check: #linebreak()  #linebreak() 
#figure(figph[Two algebraic problems demonstrate the verification of solutions for the equation 3x^-2 - 7x^-1 + 2 = 0, with x=3 and x=1/2 both resulting in 0=0.], alt: "Two algebraic problems demonstrate the verification of solutions for the equation 3x^-2 - 7x^-1 + 2 = 0, with x=3 and x=1/2 both resulting in 0=0.", caption: none)], [], [],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve: #math.equation(block: false, alt: "8 x to the power −2 minus 10 x to the power −1 plus 3 equals 0 .")[$8 x^(−2) − 10 x^(−1) + 3 = 0 .$]

#solutionbox[
#math.equation(block: true, alt: "x equals the fraction 4 over 3 x equals 2")[$x = frac(4, 3) x = 2$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve: #math.equation(block: false, alt: "6 x to the power −2 minus 23 x to the power −1 plus 20 equals 0 .")[$6 x^(−2) − 23 x^(−1) + 20 = 0 .$]

#solutionbox[
#math.equation(block: true, alt: "x equals the fraction 2 over 5 , x equals the fraction 3 over 4")[$x = frac(2, 5) , x = frac(3, 4)$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Access this online resource for additional instruction and practice with solving quadratic equations.

- #link("https://openstax.org/l/37QuadForm4")[Solving Equations in Quadratic Form]
]


=== Key Concepts

- How to solve equations in quadratic form.
+ Identify a substitution that will put the equation in quadratic form.
+ Rewrite the equation with the substitution to put it in quadratic form.
+ Solve the quadratic equation for #emph[u].
+ Substitute the original variable back into the results, using the substitution.
+ Solve for the original variable.
+ Check the solutions.


==== Practice Makes Perfect

#strong[Solve Equations in Quadratic Form]

In the following exercises, solve.

#math.equation(block: true, alt: "x to the power 4 minus 7 x squared plus 12 equals 0")[$x^(4) − 7 x^(2) + 12 = 0$]

#solutionbox[
#math.equation(block: true, alt: "x equals plus or minus the square root of 3 , x equals plus or minus 2")[$x = ± sqrt(3) , x = ± 2$]
]

#math.equation(block: true, alt: "x to the power 4 minus 9 x squared plus 18 equals 0")[$x^(4) − 9 x^(2) + 18 = 0$]

#math.equation(block: true, alt: "x to the power 4 minus 13 x squared minus 30 equals 0")[$x^(4) − 13 x^(2) − 30 = 0$]

#solutionbox[
#math.equation(block: true, alt: "x equals plus or minus the square root of 15 , x equals plus or minus the square root of 2 i")[$x = ± sqrt(15) , x = ± sqrt(2) i$]
]

#math.equation(block: true, alt: "x to the power 4 plus 5 x squared minus 36 equals 0")[$x^(4) + 5 x^(2) − 36 = 0$]

#math.equation(block: true, alt: "2 x to the power 4 minus 5 x squared plus 3 equals 0")[$2 x^(4) − 5 x^(2) + 3 = 0$]

#solutionbox[
#math.equation(block: true, alt: "x equals plus or minus 1 , x equals the fraction plus or minus the square root of 6 over 2")[$x = ± 1 , x = frac(± sqrt(6), 2)$]
]

#math.equation(block: true, alt: "4 x to the power 4 minus 5 x squared plus 1 equals 0")[$4 x^(4) − 5 x^(2) + 1 = 0$]

#math.equation(block: true, alt: "2 x to the power 4 minus 7 x squared plus 3 equals 0")[$2 x^(4) − 7 x^(2) + 3 = 0$]

#solutionbox[
#math.equation(block: true, alt: "x equals plus or minus the square root of 3 , x equals plus or minus the fraction the square root of 2 over 2")[$x = ± sqrt(3) , x = ± frac(sqrt(2), 2)$]
]

#math.equation(block: true, alt: "3 x to the power 4 minus 14 x squared plus 8 equals 0")[$3 x^(4) − 14 x^(2) + 8 = 0$]

#math.equation(block: true, alt: "open parenthesis x minus 3 close parenthesis squared minus 5 open parenthesis x minus 3 close parenthesis minus 36 equals 0")[$attach(( x − 3 ), t: 2) − 5 ( x − 3 ) − 36 = 0$]

#solutionbox[
#math.equation(block: true, alt: "x equals −1 , x equals 12")[$x = −1 , x = 12$]
]

#math.equation(block: true, alt: "open parenthesis x plus 2 close parenthesis squared minus 3 open parenthesis x plus 2 close parenthesis minus 54 equals 0")[$attach(( x + 2 ), t: 2) − 3 ( x + 2 ) − 54 = 0$]

#math.equation(block: true, alt: "open parenthesis 3 y plus 2 close parenthesis squared plus open parenthesis 3 y plus 2 close parenthesis minus 6 equals 0")[$attach(( 3 y + 2 ), t: 2) + ( 3 y + 2 ) − 6 = 0$]

#solutionbox[
#math.equation(block: true, alt: "x equals minus the fraction 5 over 3 , x equals 0")[$x = − frac(5, 3) , x = 0$]
]

#math.equation(block: true, alt: "open parenthesis 5 y minus 1 close parenthesis squared plus 3 open parenthesis 5 y minus 1 close parenthesis minus 28 equals 0")[$attach(( 5 y − 1 ), t: 2) + 3 ( 5 y − 1 ) − 28 = 0$]

#math.equation(block: true, alt: "open parenthesis x squared plus 1 close parenthesis squared minus 5 open parenthesis x squared plus 1 close parenthesis plus 4 equals 0")[$attach(( x^(2) + 1 ), t: 2) − 5 ( x^(2) + 1 ) + 4 = 0$]

#solutionbox[
#math.equation(block: true, alt: "x equals 0 , x equals plus or minus the square root of 3")[$x = 0 , x = ± sqrt(3)$]
]

#math.equation(block: true, alt: "open parenthesis x squared minus 4 close parenthesis squared minus 4 open parenthesis x squared minus 4 close parenthesis plus 3 equals 0")[$attach(( x^(2) − 4 ), t: 2) − 4 ( x^(2) − 4 ) + 3 = 0$]

#math.equation(block: true, alt: "2 open parenthesis x squared minus 5 close parenthesis squared minus 5 open parenthesis x squared minus 5 close parenthesis plus 2 equals 0")[$2 attach(( x^(2) − 5 ), t: 2) − 5 ( x^(2) − 5 ) + 2 = 0$]

#solutionbox[
#math.equation(block: true, alt: "x equals plus or minus the fraction the square root of 22 over 2 , x equals plus or minus the square root of 7")[$x = ± frac(sqrt(22), 2) , x = ± sqrt(7)$]
]

#math.equation(block: true, alt: "2 open parenthesis x squared minus 5 close parenthesis squared minus 7 open parenthesis x squared minus 5 close parenthesis plus 6 equals 0")[$2 attach(( x^(2) − 5 ), t: 2) − 7 ( x^(2) − 5 ) + 6 = 0$]

#math.equation(block: true, alt: "x minus the square root of x minus 20 equals 0")[$x − sqrt(x) − 20 = 0$]

#solutionbox[
#math.equation(block: true, alt: "x equals 25")[$x = 25$]
]

#math.equation(block: true, alt: "x minus 8 the square root of x plus 15 equals 0")[$x − 8 sqrt(x) + 15 = 0$]

#math.equation(block: true, alt: "x plus 6 the square root of x minus 16 equals 0")[$x + 6 sqrt(x) − 16 = 0$]

#solutionbox[
#math.equation(block: true, alt: "x equals 4")[$x = 4$]
]

#math.equation(block: true, alt: "x plus 4 the square root of x minus 21 equals 0")[$x + 4 sqrt(x) − 21 = 0$]

#math.equation(block: true, alt: "6 x plus the square root of x minus 2 equals 0")[$6 x + sqrt(x) − 2 = 0$]

#solutionbox[
#math.equation(block: true, alt: "x equals the fraction 1 over 4")[$x = frac(1, 4)$]
]

#math.equation(block: true, alt: "6 x plus the square root of x minus 1 equals 0")[$6 x + sqrt(x) − 1 = 0$]

#math.equation(block: true, alt: "10 x minus 17 the square root of x plus 3 equals 0")[$10 x − 17 sqrt(x) + 3 = 0$]

#solutionbox[
#math.equation(block: true, alt: "x equals the fraction 1 over 25 , x equals the fraction 9 over 4")[$x = frac(1, 25) , x = frac(9, 4)$]
]

#math.equation(block: true, alt: "12 x plus 5 the square root of x minus 3 equals 0")[$12 x + 5 sqrt(x) − 3 = 0$]

#math.equation(block: true, alt: "x to the power the fraction 2 over 3 plus 9 x to the power the fraction 1 over 3 plus 8 equals 0")[$x^(frac(2, 3)) + 9 x^(frac(1, 3)) + 8 = 0$]

#solutionbox[
#math.equation(block: true, alt: "x equals −1 , x equals −512")[$x = −1 , x = −512$]
]

#math.equation(block: true, alt: "x to the power the fraction 2 over 3 minus 3 x to the power the fraction 1 over 3 equals 28")[$x^(frac(2, 3)) − 3 x^(frac(1, 3)) = 28$]

#math.equation(block: true, alt: "x to the power the fraction 2 over 3 plus 4 x to the power the fraction 1 over 3 equals 12")[$x^(frac(2, 3)) + 4 x^(frac(1, 3)) = 12$]

#solutionbox[
#math.equation(block: true, alt: "x equals 8 , x equals −216")[$x = 8 , x = −216$]
]

#math.equation(block: true, alt: "x to the power the fraction 2 over 3 minus 11 x to the power the fraction 1 over 3 plus 30 equals 0")[$x^(frac(2, 3)) − 11 x^(frac(1, 3)) + 30 = 0$]

#math.equation(block: true, alt: "6 x to the power the fraction 2 over 3 minus x to the power the fraction 1 over 3 equals 12")[$6 x^(frac(2, 3)) − x^(frac(1, 3)) = 12$]

#solutionbox[
#math.equation(block: true, alt: "x equals the fraction 27 over 8 , x equals minus the fraction 64 over 27")[$x = frac(27, 8) , x = − frac(64, 27)$]
]

#math.equation(block: true, alt: "3 x to the power the fraction 2 over 3 minus 10 x to the power the fraction 1 over 3 equals 8")[$3 x^(frac(2, 3)) − 10 x^(frac(1, 3)) = 8$]

#math.equation(block: true, alt: "8 x to the power the fraction 2 over 3 minus 43 x to the power the fraction 1 over 3 plus 15 equals 0")[$8 x^(frac(2, 3)) − 43 x^(frac(1, 3)) + 15 = 0$]

#solutionbox[
#math.equation(block: true, alt: "x equals the fraction 27 over 512 , x equals 125")[$x = frac(27, 512) , x = 125$]
]

#math.equation(block: true, alt: "20 x to the power the fraction 2 over 3 minus 23 x to the power the fraction 1 over 3 plus 6 equals 0")[$20 x^(frac(2, 3)) − 23 x^(frac(1, 3)) + 6 = 0$]

#math.equation(block: true, alt: "x minus 8 x to the power the fraction 1 over 2 plus 7 equals 0")[$x − 8 x^(frac(1, 2)) + 7 = 0$]

#solutionbox[
#math.equation(block: true, alt: "x equals 1 , x equals 49")[$x = 1 , x = 49$]
]

#math.equation(block: true, alt: "2 x minus 7 x to the power the fraction 1 over 2 equals 15")[$2 x − 7 x^(frac(1, 2)) = 15$]

#math.equation(block: true, alt: "6 x to the power −2 plus 13 x to the power −1 plus 5 equals 0")[$6 x^(−2) + 13 x^(−1) + 5 = 0$]

#solutionbox[
#math.equation(block: true, alt: "x equals −2 , x equals minus the fraction 3 over 5")[$x = −2 , x = − frac(3, 5)$]
]

#math.equation(block: true, alt: "15 x to the power −2 minus 26 x to the power −1 plus 8 equals 0")[$15 x^(−2) − 26 x^(−1) + 8 = 0$]

#math.equation(block: true, alt: "8 x to the power −2 minus 2 x to the power −1 minus 3 equals 0")[$8 x^(−2) − 2 x^(−1) − 3 = 0$]

#solutionbox[
#math.equation(block: true, alt: "x equals −2 , x equals the fraction 4 over 3")[$x = −2 , x = frac(4, 3)$]
]

#math.equation(block: true, alt: "15 x to the power −2 minus 4 x to the power −1 minus 4 equals 0")[$15 x^(−2) − 4 x^(−1) − 4 = 0$]


==== Writing Exercises

Explain how to recognize an equation in quadratic form.

#solutionbox[
Answers will vary.
]

Explain the procedure for solving an equation in quadratic form.


==== Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

#figure(figph[This table provides a checklist to evaluate mastery of the objectives of this section. Choose how would you respond to the statement “I can solve equations in quadratic form.” “Confidently,” “with some help,” or “No, I don’t get it.”], alt: "This table provides a checklist to evaluate mastery of the objectives of this section. Choose how would you respond to the statement “I can solve equations in quadratic form.” “Confidently,” “with some help,” or “No, I don’t get it.”", caption: none)

ⓑ On a scale of 1-10, how would you rate your mastery of this section in light of your responses on the checklist? How can you improve this?