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== 9.1#h(0.6em)Solve Quadratic Equations Using the Square Root Property

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

Simplify: #math.equation(block: false, alt: "the square root of 128 .")[$sqrt(128) .$] #linebreak() 
If you missed this problem, review .

#solutionbox[
#math.equation(block: true, alt: "8 the square root of 2")[$8 sqrt(2)$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Simplify: #math.equation(block: false, alt: "the square root of the fraction 32 over 5")[$sqrt(frac(32, 5))$]. #linebreak() 
If you missed this problem, review .

#solutionbox[
#math.equation(block: true, alt: "the fraction 4 the square root of 10 over 5")[$frac(4 sqrt(10), 5)$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Factor: #math.equation(block: false, alt: "9 x squared minus 12 x plus 4")[$9 x^(2) − 12 x + 4$]. #linebreak() 
If you missed this problem, review .

#solutionbox[
#math.equation(block: true, alt: "3 x minus 2 squared")[$attach(3 x − 2, t: 2)$]
]
]

A #strong[quadratic equation] is an equation of the form #emph[ax]#super[2] + #emph[bx] + #emph[c] = 0, where #math.equation(block: false, alt: "a not equal to 0")[$a ≠ 0$]. Quadratic equations differ from linear equations by including a quadratic term with the variable raised to the second power of the form #emph[ax]#super[2]. We use different methods to solve quadratic equations than linear equations, because just adding, subtracting, multiplying, and dividing terms will not isolate the variable.

We have seen that some quadratic equations can be solved by factoring. In this chapter, we will learn three other methods to use in case a quadratic equation cannot be factored.


=== Solve Quadratic Equations of the form #math.equation(block: false, alt: "a x squared equals k")[$a x^(2) = k$] using the Square Root Property

We have already solved some quadratic equations by factoring. Let’s review how we used factoring to solve the quadratic equation #emph[x]#super[2] = 9.

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#math.equation(block: false, alt: "x squared equals 9")[$x^(2) = 9 #h(4em)$]]),
  [Put the equation in standard form.], [#math.equation(block: false, alt: "x squared minus 9 equals 0")[$x^(2) − 9 = 0 #h(4em)$]],
  [Factor the difference of squares.], [#math.equation(block: false, alt: "open parenthesis x minus 3 close parenthesis open parenthesis x plus 3 close parenthesis equals 0")[$( x − 3 ) ( x + 3 ) = 0 #h(4em)$]],
  [Use the Zero Product Property.], [#math.equation(block: false, alt: "x minus 3 equals 0 x minus 3 equals 0")[$x − 3 = 0 #h(2em) x − 3 = 0 #h(2em)$]],
  [Solve each equation.], [#math.equation(block: false, alt: "x equals 3 x equals −3")[$x = 3 #h(4em) x = −3 #h(1.3em)$]],
))

We can easily use factoring to find the solutions of similar equations, like #emph[x]#super[2] = 16 and #emph[x]#super[2] = 25, because 16 and 25 are perfect squares. In each case, we would get two solutions, #math.equation(block: false, alt: "x equals 4 , x equals −4")[$x = 4 , x = −4$] and #math.equation(block: false, alt: "x equals 5 , x equals −5 .")[$x = 5 , x = −5 .$]

But what happens when we have an equation like #emph[x]#super[2] = 7? Since 7 is not a perfect square, we cannot solve the equation by factoring.

Previously we learned that since 169 is the square of 13, we can also say that 13 is a #emph[square root] of 169. Also, (−13)#super[2] = 169, so −13 is also a square root of 169. Therefore, both 13 and −13 are square roots of 169. So, every positive number has two square roots—one positive and one negative. We earlier defined the square root of a number in this way:

#math.equation(block: true, alt: "If n squared equals m , then n is a square root of m .")[$"If" #h(0.2em) n^(2) = m , #h(0.2em) "then" #h(0.2em) n #h(0.2em) "is a square root of" #h(0.2em) m .$]Since these equations are all of the form #emph[x]#super[2] = #emph[k], the square root definition tells us the solutions are the two square roots of #emph[k]. This leads to the #strong[Square Root Property].

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Square Root Property]

If #emph[x]#super[2] = #emph[k], then

#math.equation(block: true, alt: "x equals the square root of k or x equals − the square root of k or x equals plus or minus the square root of k .")[$x = sqrt(k) #h(1em) "or" #h(1em) x = "−" sqrt(k) #h(1em) "or" #h(1em) x = ± sqrt(k) .$]
]

Notice that the Square Root Property gives two solutions to an equation of the form #emph[x]#super[2] = #emph[k], the principal square root of #math.equation(block: false, alt: "k")[$k$] and its opposite. We could also write the solution as #math.equation(block: false, alt: "x equals plus or minus the square root of k .")[$x = ± sqrt(k) .$] We read this as #emph[x] equals positive or negative the square root of #emph[k].

Now we will solve the equation #emph[x]#super[2] = 9 again, this time using the Square Root Property.

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#math.equation(block: false, alt: "x squared equals 9")[$#h(2em) x^(2) = 9$]]),
  [Use the Square Root Property.], [#math.equation(block: false, alt: "x equals plus or minus the square root of 9")[$#h(2.5em) x = ± sqrt(9)$]],
  [], [#math.equation(block: false, alt: "x equals plus or minus 3")[$#h(2.5em) x = ± 3$]],
  [], [#math.equation(block: false, alt: "So x equals 3 or x equals −3 .")[$"So" #h(0.5em) x = 3 #h(0.5em) "or" #h(0.5em) x = −3 .$]],
))

What happens when the constant is not a perfect square? Let’s use the Square Root Property to solve the equation #emph[x]#super[2] = 7.

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#math.equation(block: false, alt: "x squared equals 7")[$x^(2) = 7$]]),
  [Use the Square Root Property.], [#math.equation(block: false, alt: "x equals the square root of 7 , x equals − the square root of 7")[$x = sqrt(7) , #h(1em) x = "−" sqrt(7)$]],
))

We cannot simplify #math.equation(block: false, alt: "the square root of 7")[$sqrt(7)$], so we leave the answer as a radical.

#examplebox("Example 1")[How to solve a Quadratic Equation of the form #emph[ax]#super[2] = #emph[k] Using the Square Root Property][
Solve: #math.equation(block: false, alt: "x squared minus 50 equals 0 .")[$x^(2) − 50 = 0 .$]

#solutionbox[
#figure(figph[Step one is to isolate the quadratic term and make its coefficient one. For the equation x squared minus fifty equals zero, first add fifty to both sides to get x squared by itself. The new equation is x squared equals fifty.], alt: "Step one is to isolate the quadratic term and make its coefficient one. For the equation x squared minus fifty equals zero, first add fifty to both sides to get x squared by itself. The new equation is x squared equals fifty.", caption: none)

#figure(figph[Step two is to use Square Root Property. Remember to write the plus or minus symbol. The equation created is x equals the positive or negative square root of 50.], alt: "Step two is to use Square Root Property. Remember to write the plus or minus symbol. The equation created is x equals the positive or negative square root of 50.", caption: none)

#figure(figph[Step three is to simplify the radical if possible. Continue to write equivalent equations. X equals the positive or negative square root of twenty five times the square root of five. X equals positive or negative five times the square root of five. Rewrite to show two solutions: x equals five times the square root of five or x equals negative five times the square root of five.], alt: "Step three is to simplify the radical if possible. Continue to write equivalent equations. X equals the positive or negative square root of twenty five times the square root of five. X equals positive or negative five times the square root of five. Rewrite to show two solutions: x equals five times the square root of five or x equals negative five times the square root of five.", caption: none)

#figure(figph[Checking solutions for a quadratic equation. Substitute x equals 5 times the square root of 5 and x equals negative 5 times the square root of 5 into the original equation to verify both are valid solutions.], alt: "Checking solutions for a quadratic equation. Substitute x equals 5 times the square root of 5 and x equals negative 5 times the square root of 5 into the original equation to verify both are valid solutions.", caption: none)
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve: #math.equation(block: false, alt: "x squared minus 48 equals 0 .")[$x^(2) − 48 = 0 .$]

#solutionbox[
#math.equation(block: true, alt: "x equals 4 the square root of 3 , x equals −4 the square root of 3")[$x = 4 sqrt(3) , x = −4 sqrt(3)$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve: #math.equation(block: false, alt: "y squared minus 27 equals 0 .")[$y^(2) − 27 = 0 .$]

#solutionbox[
#math.equation(block: true, alt: "y equals 3 the square root of 3 , y equals −3 the square root of 3")[$y = 3 sqrt(3) , y = −3 sqrt(3)$]
]
]

The steps to take to use the #strong[Square Root Property] to solve a quadratic equation are listed here.

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Solve a quadratic equation using the square root property.]

+ Isolate the quadratic term and make its coefficient one.
+ Use Square Root Property.
+ Simplify the radical.
+ Check the solutions.
]

In order to use the Square Root Property, the coefficient of the variable term must equal one. In the next example, we must divide both sides of the equation by the coefficient 3 before using the Square Root Property.

#examplebox("Example 2")[][
Solve: #math.equation(block: false, alt: "3 z squared equals 108 .")[$3 z^(2) = 108 .$]

#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#math.equation(block: false, alt: "3 z squared equals 108")[$3 z^(2) = 108$]]),
  [The quadratic term is isolated. #linebreak() 
Divide by 3 to make its coefficient 1.], [#math.equation(block: false, alt: "the fraction 3 z squared over 3 equals the fraction 108 over 3")[$frac(3 z^(2), 3) = frac(108, 3)$]],
  [Simplify.], [#math.equation(block: false, alt: "z squared equals 36")[$z^(2) = 36$]],
  [Use the Square Root Property.], [#math.equation(block: false, alt: "z equals plus or minus the square root of 36")[$#h(2em) z = ± sqrt(36)$]],
  [Simplify the radical.], [#math.equation(block: false, alt: "z equals plus or minus 6")[$#h(1.1em) z = ± 6$]],
  [Rewrite to show two solutions.], [#math.equation(block: false, alt: "z equals 6 , z equals −6")[$z = 6 , #h(0.5em) z = −6$]],
  [Check the solutions: #linebreak()  #linebreak() 
#figure(figph[This image demonstrates how to verify the solutions z = 6 and z = -6 for the equation 3z^2 = 108, showing that both positive and negative square roots satisfy the equation.], alt: "This image demonstrates how to verify the solutions z = 6 and z = -6 for the equation 3z^2 = 108, showing that both positive and negative square roots satisfy the equation.", caption: none)], [],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve: #math.equation(block: false, alt: "2 x squared equals 98 .")[$2 x^(2) = 98 .$]

#solutionbox[
#math.equation(block: true, alt: "x equals 7 , x equals −7")[$x = 7 , x = −7$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve: #math.equation(block: false, alt: "5 m squared equals 80 .")[$5 m^(2) = 80 .$]

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

The Square Root Property states ‘If #math.equation(block: false, alt: "x squared equals k")[$x^(2) = k$],’ What will happen if #math.equation(block: false, alt: "k less than 0 ?")[$k < 0 ?$] This will be the case in the next example.

#examplebox("Example 3")[][
Solve: #math.equation(block: false, alt: "x squared plus 72 equals 0")[$x^(2) + 72 = 0$].

#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#math.equation(block: false, alt: "x squared plus 72 equals 0")[$x^(2) + 72 = 0$]]),
  [Isolate the quadratic term.], [#math.equation(block: false, alt: "x squared equals −72")[$#h(2.15em) x^(2) = −72$]],
  [Use the Square Root Property.], [#math.equation(block: false, alt: "x equals plus or minus the square root of −72")[$#h(2.6em) x = ± sqrt(−72)$]],
  [Simplify using complex numbers.], [#math.equation(block: false, alt: "x equals plus or minus the square root of 72 i")[$#h(2.6em) x = ± sqrt(72) #h(0.2em) i$]],
  [Simplify the radical.], [#math.equation(block: false, alt: "x equals plus or minus 6 the square root of 2 i")[$#h(2.6em) x = ± 6 sqrt(2) #h(0.2em) i$]],
  [Rewrite to show two solutions.], [#math.equation(block: false, alt: "x equals 6 the square root of 2 i , x equals −6 the square root of 2 i")[$x = 6 sqrt(2) #h(0.2em) i , #h(0.5em) x = −6 sqrt(2) #h(0.2em) i$]],
  [Check the solutions: #linebreak()  #linebreak() 
#figure(figph[This image verifies that both x = 6√2i and x = -6√2i are valid solutions for the quadratic equation x^2 + 72 = 0 through step-by-step substitution.], alt: "This image verifies that both x = 6√2i and x = -6√2i are valid solutions for the quadratic equation x^2 + 72 = 0 through step-by-step substitution.", caption: none)], [],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve: #math.equation(block: false, alt: "c squared plus 12 equals 0 .")[$c^(2) + 12 = 0 .$]

#solutionbox[
#math.equation(block: true, alt: "c equals 2 the square root of 3 i , c equals −2 the square root of 3 i")[$c = 2 sqrt(3) i , #h(0.2em) #h(0.2em) c = −2 sqrt(3) i$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve: #math.equation(block: false, alt: "q squared plus 24 equals 0 .")[$q^(2) + 24 = 0 .$]

#solutionbox[
#math.equation(block: true, alt: "c equals 2 the square root of 6 i , c equals −2 the square root of 6 i")[$c = 2 sqrt(6) i , #h(0.2em) #h(0.2em) c = −2 sqrt(6) i$]
]
]

Our method also works when fractions occur in the equation; we solve as any equation with fractions. In the next example, we first isolate the quadratic term, and then make the coefficient equal to one.

#examplebox("Example 4")[][
Solve: #math.equation(block: false, alt: "the fraction 2 over 3 u squared plus 5 equals 17 .")[$frac(2, 3) u^(2) + 5 = 17 .$]

#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#math.equation(block: false, alt: "the fraction 2 over 3 u squared plus 5 equals 17")[$frac(2, 3) u^(2) + 5 = 17$]]),
  [Isolate the quadratic term.], [#figure(figph[A mathematical equation is displayed on a white background, which reads '2/3 u^2 = 12'.], alt: "A mathematical equation is displayed on a white background, which reads '2/3 u^2 = 12'.", caption: none)],
  [Multiply by #math.equation(block: false, alt: "the fraction 3 over 2")[$frac(3, 2)$] to make the coefficient 1.], [#figure(figph[An algebraic equation: (3/2) multiplied by (2/3)u squared equals (3/2) multiplied by 12, with the fraction 3/2 highlighted in red on both sides.], alt: "An algebraic equation: (3/2) multiplied by (2/3)u squared equals (3/2) multiplied by 12, with the fraction 3/2 highlighted in red on both sides.", caption: none)],
  [Simplify.], [#figure(figph[A mathematical equation is displayed, showing u squared equals eighteen (u^2 = 18). The characters are in a dark grey font against a plain white background.], alt: "A mathematical equation is displayed, showing u squared equals eighteen (u^2 = 18). The characters are in a dark grey font against a plain white background.", caption: none)],
  [Use the Square Root Property.], [#figure(figph[A mathematical expression displays u equals plus or minus the square root of 18.], alt: "A mathematical expression displays u equals plus or minus the square root of 18.", caption: none)],
  [Simplify the radical.], [#figure(figph[A mathematical equation shows 'u = plus-minus square root of 9 multiplied by 2' against a white background.], alt: "A mathematical equation shows 'u = plus-minus square root of 9 multiplied by 2' against a white background.", caption: none)],
  [Simplify.], [#figure(figph[The image shows the equation u = ±3√2, representing two possible values for u: positive three times the square root of two, and negative three times the square root of two.], alt: "The image shows the equation u = ±3√2, representing two possible values for u: positive three times the square root of two, and negative three times the square root of two.", caption: none)],
  [Rewrite to show two solutions.], [#figure(figph[Two solutions for 'u' are presented: u = 3√2 and u = -3√2, representing positive and negative square roots in a mathematical context.], alt: "Two solutions for 'u' are presented: u = 3√2 and u = -3√2, representing positive and negative square roots in a mathematical context.", caption: none)],
  [Check: #linebreak()  #linebreak() 
#figure(figph[Verification of two solutions, u = 3√2 and u = -3√2, for the equation (2/3)u^2 + 5 = 17, showing step-by-step calculations that confirm both values satisfy the equation.], alt: "Verification of two solutions, u = 3√2 and u = -3√2, for the equation (2/3)u^2 + 5 = 17, showing step-by-step calculations that confirm both values satisfy the equation.", caption: none)], [],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve: #math.equation(block: false, alt: "the fraction 1 over 2 x squared plus 4 equals 24 .")[$frac(1, 2) x^(2) + 4 = 24 .$]

#solutionbox[
#math.equation(block: true, alt: "x equals 2 the square root of 10 , x equals −2 the square root of 10")[$x = 2 sqrt(10) , #h(0.5em) x = −2 sqrt(10)$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve: #math.equation(block: false, alt: "the fraction 3 over 4 y squared minus 3 equals 18 .")[$frac(3, 4) y^(2) − 3 = 18 .$]

#solutionbox[
#math.equation(block: true, alt: "y equals 2 the square root of 7 , y equals −2 the square root of 7")[$y = 2 sqrt(7) , #h(0.5em) y = −2 sqrt(7)$]
]
]

The solutions to some equations may have fractions inside the radicals. When this happens, we must #strong[rationalize the denominator].

#examplebox("Example 5")[][
Solve: #math.equation(block: false, alt: "2 x squared minus 8 equals 41 .")[$2 x^(2) − 8 = 41 .$]

#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#figure(figph[A mathematical equation is displayed, reading '2x^2 - 8 = 41' on a white background.], alt: "A mathematical equation is displayed, reading '2x^2 - 8 = 41' on a white background.", caption: none)]),
  [Isolate the quadratic term.], [#figure(figph[The image displays the mathematical equation 2x^2 = 49, presented in a clean, straightforward manner against a white background.], alt: "The image displays the mathematical equation 2x^2 = 49, presented in a clean, straightforward manner against a white background.", caption: none)],
  [Divide by #math.equation(block: false, alt: "2")[$2$] to make the coefficient 1.], [#figure(figph[A mathematical equation is displayed, showing '2x^2 / 2 = 49 / 2' in black text against a white background.], alt: "A mathematical equation is displayed, showing '2x^2 / 2 = 49 / 2' in black text against a white background.", caption: none)],
  [Simplify.], [#figure(figph[A mathematical equation displays 'X squared equals 49 over 2' on a white background.], alt: "A mathematical equation displays 'X squared equals 49 over 2' on a white background.", caption: none)],
  [Use the Square Root Property.], [#figure(figph[The mathematical equation 'x = ××±×× sqrt(49/2)' is displayed in black text on a white background.], alt: "The mathematical equation 'x = ××±×× sqrt(49/2)' is displayed in black text on a white background.", caption: none)],
  [Rewrite the radical as a fraction of square roots.], [#figure(figph[A mathematical equation shows x equals plus or minus the square root of 49 divided by the square root of 2.], alt: "A mathematical equation shows x equals plus or minus the square root of 49 divided by the square root of 2.", caption: none)],
  [Rationalize the denominator.], [#figure(figph[A mathematical equation shows x equals plus or minus a fraction where the numerator is the square root of 49 multiplied by the square root of 2, and the denominator is the square root of 2 multiplied by the square root of 2. The second √2 in both the numerator and denominator is in red.], alt: "A mathematical equation shows x equals plus or minus a fraction where the numerator is the square root of 49 multiplied by the square root of 2, and the denominator is the square root of 2 multiplied by the square root of 2. The second √2 in both the numerator and denominator is in red.", caption: none)],
  [Simplify.], [#figure(figph[A mathematical equation shows x equals plus or minus seven times the square root of two, all divided by two.], alt: "A mathematical equation shows x equals plus or minus seven times the square root of two, all divided by two.", caption: none)],
  [Rewrite to show two solutions.], [#figure(figph[The image displays two solutions for 'x': x = 7√2 / 2 and x = -7√2 / 2.], alt: "The image displays two solutions for 'x': x = 7√2 / 2 and x = -7√2 / 2.", caption: none)],
  [Check: #linebreak() We leave the check for you.], [],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve: #math.equation(block: false, alt: "5 r squared minus 2 equals 34 .")[$5 r^(2) − 2 = 34 .$]

#solutionbox[
#math.equation(block: true, alt: "r equals the fraction 6 the square root of 5 over 5 , r equals minus the fraction 6 the square root of 5 over 5")[$r = frac(6 sqrt(5), 5) , #h(0.5em) r = − frac(6 sqrt(5), 5)$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve: #math.equation(block: false, alt: "3 t squared plus 6 equals 70 .")[$3 t^(2) + 6 = 70 .$]

#solutionbox[
#math.equation(block: true, alt: "t equals the fraction 8 the square root of 3 over 3 , t equals minus the fraction 8 the square root of 3 over 3")[$t = frac(8 sqrt(3), 3) , #h(0.5em) t = − frac(8 sqrt(3), 3)$]
]
]


=== Solve Quadratic Equations of the Form #emph[a] (#emph[x] − #emph[h])#super[2] = #emph[k] Using the Square Root Property

We can use the #strong[Square Root Property] to solve an equation of the form #emph[a] (#emph[x] − #emph[h])#super[2] = #emph[k] as well. Notice that the quadratic term, #emph[x], in the original form #emph[ax]#super[2] = #emph[k] is replaced with (#emph[x] − #emph[h]).

#figure(figph[On the left is the equation a times x square equals k. Replacing x in this equation with the expression x minus h changes the equation. It is now a times the square of x minus h equals k.], alt: "On the left is the equation a times x square equals k. Replacing x in this equation with the expression x minus h changes the equation. It is now a times the square of x minus h equals k.", caption: none)

The first step, like before, is to isolate the term that has the variable squared. In this case, a binomial is being squared. Once the binomial is isolated, by dividing each side by the coefficient of #emph[a], then the Square Root Property can be used on (#emph[x] − #emph[h])#super[2].

#examplebox("Example 6")[][
Solve: #math.equation(block: false, alt: "4 open parenthesis y minus 7 close parenthesis squared equals 48 .")[$4 attach(( y − 7 ), t: 2) = 48 .$]

#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  [], [#math.equation(block: false, alt: "4 open parenthesis y minus 7 close parenthesis squared equals 48")[$4 attach(( y − 7 ), t: 2) = 48$]],
  [Divide both sides by the coefficient 4.], [#math.equation(block: false, alt: "open parenthesis y minus 7 close parenthesis squared equals 12")[$#h(0.5em) attach(( y − 7 ), t: 2) = 12$]],
  [Use the Square Root Property on the binomial], [#math.equation(block: false, alt: "y minus 7 equals plus or minus the square root of 12")[$#h(1.3em) y − 7 = ± sqrt(12)$]],
  [Simplify the radical.], [#math.equation(block: false, alt: "y minus 7 equals plus or minus 2 the square root of 3")[$#h(1.3em) y − 7 = ± 2 sqrt(3)$]],
  [Solve for #math.equation(block: false, alt: "y .")[$y .$]], [#math.equation(block: false, alt: "y equals 7 plus or minus 2 the square root of 3")[$#h(3em) y = 7 ± 2 sqrt(3)$]],
  [Rewrite to show two solutions.], [#math.equation(block: false, alt: "y equals 7 plus 2 the square root of 3 ,")[$y = 7 + 2 sqrt(3) ,$] #math.equation(block: false, alt: "y equals 7 minus 2 the square root of 3")[$y = 7 − 2 sqrt(3)$]],
  [Check: #linebreak()  #linebreak() 
#figure(figph[Step-by-step verification of the solutions y = 7 + 2√3 and y = 7 - 2√3 for the equation 4(y-7)^2 = 48, showing that both satisfy the equation and result in 48=48.], alt: "Step-by-step verification of the solutions y = 7 + 2√3 and y = 7 - 2√3 for the equation 4(y-7)^2 = 48, showing that both satisfy the equation and result in 48=48.", caption: none)], [],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve: #math.equation(block: false, alt: "3 open parenthesis a minus 3 close parenthesis squared equals 54 .")[$3 attach(( a − 3 ), t: 2) = 54 .$]

#solutionbox[
#math.equation(block: true, alt: "a equals 3 plus 3 the square root of 2 , a equals 3 minus 3 the square root of 2")[$a = 3 + 3 sqrt(2) , #h(0.5em) a = 3 − 3 sqrt(2)$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve: #math.equation(block: false, alt: "2 open parenthesis b plus 2 close parenthesis squared equals 80 .")[$2 attach(( b + 2 ), t: 2) = 80 .$]

#solutionbox[
#math.equation(block: true, alt: "b equals −2 plus 2 the square root of 10 , b equals −2 minus 2 the square root of 10")[$b = −2 + 2 sqrt(10) , #h(0.5em) b = −2 − 2 sqrt(10)$]
]
]

Remember when we take the square root of a fraction, we can take the square root of the numerator and denominator separately.

#examplebox("Example 7")[][
Solve: #math.equation(block: false, alt: "open parenthesis x minus the fraction 1 over 3 close parenthesis squared equals the fraction 5 over 9 .")[$attach(( x − frac(1, 3) ), t: 2) = frac(5, 9) .$]

#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#math.equation(block: false, alt: "open parenthesis x minus the fraction 1 over 3 close parenthesis squared equals the fraction 5 over 9")[$#h(0.5em) attach(( x − frac(1, 3) ), t: 2) = frac(5, 9)$]]),
  [Use the Square Root Property.], [#math.equation(block: false, alt: "x minus the fraction 1 over 3 equals plus or minus the square root of the fraction 5 over 9")[$#h(2.2em) x − frac(1, 3) = ± sqrt(frac(5, 9))$]],
  [Rewrite the radical as a fraction of square roots.], [#math.equation(block: false, alt: "x minus the fraction 1 over 3 equals plus or minus the fraction the square root of 5 over the square root of 9")[$#h(2.2em) x − frac(1, 3) = ± frac(sqrt(5), sqrt(9))$]],
  [Simplify the radical.], [#math.equation(block: false, alt: "x minus the fraction 1 over 3 equals plus or minus the fraction the square root of 5 over 3")[$#h(2.2em) x − frac(1, 3) = ± frac(sqrt(5), 3)$]],
  [Solve for #math.equation(block: false, alt: "x")[$x$].], [#math.equation(block: false, alt: "x equals the fraction 1 over 3 plus or minus the fraction the square root of 5 over 3")[$#h(4.2em) x = frac(1, 3) ± frac(sqrt(5), 3)$]],
  [Rewrite to show two solutions.], [#math.equation(block: false, alt: "x equals the fraction 1 over 3 plus the fraction the square root of 5 over 3 , x equals the fraction 1 over 3 minus the fraction the square root of 5 over 3")[$x = frac(1, 3) + frac(sqrt(5), 3) , #h(0.5em) x = frac(1, 3) − frac(sqrt(5), 3)$]],
  [Check: #linebreak() 
We leave the check for you.], [],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve: #math.equation(block: false, alt: "open parenthesis x minus the fraction 1 over 2 close parenthesis squared equals the fraction 5 over 4 .")[$attach(( x − frac(1, 2) ), t: 2) = frac(5, 4) .$]

#solutionbox[
#math.equation(block: true, alt: "x equals the fraction 1 over 2 plus the fraction the square root of 5 over 2")[$x = frac(1, 2) + frac(sqrt(5), 2)$],#math.equation(block: true, alt: "x equals the fraction 1 over 2 minus the fraction the square root of 5 over 2")[$x = frac(1, 2) − frac(sqrt(5), 2)$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve: #math.equation(block: false, alt: "open parenthesis y plus the fraction 3 over 4 close parenthesis squared equals the fraction 7 over 16 .")[$attach(( y + frac(3, 4) ), t: 2) = frac(7, 16) .$]

#solutionbox[
#math.equation(block: true, alt: "y equals minus the fraction 3 over 4 plus the fraction the square root of 7 over 4 , y equals minus the fraction 3 over 4 minus the fraction the square root of 7 over 4")[$y = − frac(3, 4) + frac(sqrt(7), 4) , #h(0.5em) y = − frac(3, 4) − frac(sqrt(7), 4)$]
]
]

We will start the solution to the next example by isolating the binomial term.

#examplebox("Example 8")[][
Solve: #math.equation(block: false, alt: "2 open parenthesis x minus 2 close parenthesis squared plus 3 equals 57 .")[$2 attach(( x − 2 ), t: 2) + 3 = 57 .$]

#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#math.equation(block: false, alt: "2 open parenthesis x minus 2 close parenthesis squared plus 3 equals 57")[$2 attach(( x − 2 ), t: 2) + 3 = 57 #h(2.8em)$]]),
  [Subtract 3 from both sides to isolate the binomial term.], [#math.equation(block: false, alt: "2 open parenthesis x minus 2 close parenthesis squared equals 54")[$2 attach(( x − 2 ), t: 2) = 54 #h(2.8em)$]],
  [Divide both sides by 2.], [#math.equation(block: false, alt: "open parenthesis x minus 2 close parenthesis squared equals 27")[$attach(( x − 2 ), t: 2) = 27 #h(2.8em)$]],
  [Use the Square Root Property.], [#math.equation(block: false, alt: "x minus 2 equals plus or minus the square root of 27")[$x − 2 = ± sqrt(27) #h(1em)$]],
  [Simplify the radical.], [#math.equation(block: false, alt: "x minus 2 equals plus or minus 3 the square root of 3")[$x − 2 = ± 3 sqrt(3) #h(1em)$]],
  [Solve for #math.equation(block: false, alt: "x")[$x$].], [#math.equation(block: false, alt: "x equals 2 plus or minus 3 the square root of 3")[$x = 2 ± 3 sqrt(3) #h(0.1em)$]],
  [Rewrite to show two solutions.], [#math.equation(block: false, alt: "x equals 2 plus 3 the square root of 3 , x equals 2 minus 3 the square root of 3")[$x = 2 + 3 sqrt(3) , #h(0.5em) x = 2 − 3 sqrt(3)$]],
  [Check: #linebreak() 
We leave the check for you.], [],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve: #math.equation(block: false, alt: "5 open parenthesis a minus 5 close parenthesis squared plus 4 equals 104 .")[$5 attach(( a − 5 ), t: 2) + 4 = 104 .$]

#solutionbox[
#math.equation(block: true, alt: "a equals 5 plus 2 the square root of 5 , a equals 5 minus 2 the square root of 5")[$a = 5 + 2 sqrt(5) , #h(0.5em) a = 5 − 2 sqrt(5)$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve: #math.equation(block: false, alt: "3 open parenthesis b plus 3 close parenthesis squared minus 8 equals 88 .")[$3 attach(( b + 3 ), t: 2) − 8 = 88 .$]

#solutionbox[
#math.equation(block: true, alt: "b equals −3 plus 4 the square root of 2 , b equals −3 minus 4 the square root of 2")[$b = −3 + 4 sqrt(2) , #h(0.5em) b = −3 − 4 sqrt(2)$]
]
]

Sometimes the solutions are complex numbers.

#examplebox("Example 9")[][
Solve: #math.equation(block: false, alt: "open parenthesis 2 x minus 3 close parenthesis squared equals −12 .")[$attach(( 2 x − 3 ), t: 2) = −12 .$]

#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#math.equation(block: false, alt: "open parenthesis 2 x minus 3 close parenthesis squared equals −12")[$#h(3.2em) attach(( 2 x − 3 ), t: 2) = −12$]]),
  [Use the Square Root Property.], [#math.equation(block: false, alt: "2 x minus 3 equals plus or minus the square root of −12")[$#h(4.5em) 2 x − 3 = ± sqrt(−12)$]],
  [Simplify the radical.], [#math.equation(block: false, alt: "2 x minus 3 equals plus or minus 2 the square root of 3 i")[$#h(4.5em) 2 x − 3 = ± 2 sqrt(3) #h(0.2em) i$]],
  [Add 3 to both sides.], [#math.equation(block: false, alt: "2 x equals 3 plus or minus 2 the square root of 3 i")[$#h(6.2em) 2 x = 3 ± 2 sqrt(3) #h(0.2em) i$]],
  [Divide both sides by 2.], [#math.equation(block: false, alt: "x equals the fraction 3 plus or minus 2 the square root of 3 i over 2")[$#h(6.7em) x = frac(3 ± 2 sqrt(3) #h(0.2em) i, 2)$]],
  [Rewrite in standard form.], [#math.equation(block: false, alt: "x equals the fraction 3 over 2 plus or minus the fraction 2 the square root of 3 i over 2")[$#h(6.7em) x = frac(3, 2) ± frac(2 sqrt(3) #h(0.2em) i, 2)$]],
  [Simplify.], [#math.equation(block: false, alt: "x equals the fraction 3 over 2 plus or minus the square root of 3 i")[$#h(6.7em) x = frac(3, 2) ± sqrt(3) #h(0.2em) i$]],
  [Rewrite to show two solutions.], [#math.equation(block: false, alt: "x equals the fraction 3 over 2 plus the square root of 3 i , x equals the fraction 3 over 2 minus the square root of 3 i")[$x = frac(3, 2) + sqrt(3) #h(0.2em) i , #h(0.5em) x = frac(3, 2) − sqrt(3) #h(0.2em) i$]],
  [Check: #linebreak() 
We leave the check for you.], [],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve: #math.equation(block: false, alt: "open parenthesis 3 r plus 4 close parenthesis squared equals −8 .")[$attach(( 3 r + 4 ), t: 2) = −8 .$]

#solutionbox[
#math.equation(block: true, alt: "r equals minus the fraction 4 over 3 plus the fraction 2 the square root of 2 i over 3 , r equals minus the fraction 4 over 3 minus the fraction 2 the square root of 2 i over 3")[$r = − frac(4, 3) + frac(2 sqrt(2) i, 3) , #h(0.2em) "r" = − frac(4, 3) − frac(2 sqrt(2) i, 3)$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve: #math.equation(block: false, alt: "open parenthesis 2 t minus 8 close parenthesis squared equals −10 .")[$attach(( 2 t − 8 ), t: 2) = −10 .$]

#solutionbox[
#math.equation(block: true, alt: "t equals 4 plus the fraction the square root of 10 i over 2 , t equals 4 minus the fraction the square root of 10 i over 2")[$t = 4 + frac(sqrt(10) i, 2) , #h(0.2em) "t" = 4 − frac(sqrt(10) i, 2)$]
]
]

The left sides of the equations in the next two examples do not seem to be of the form #emph[a] (#emph[x] − #emph[h])#super[2]. But they are perfect square trinomials, so we will factor to put them in the form we need.

#examplebox("Example 10")[][
Solve: #math.equation(block: false, alt: "4 n squared plus 4 n plus 1 equals 16 .")[$4 n^(2) + 4 n + 1 = 16 .$]

#solutionbox[
We notice the left side of the equation is a perfect square trinomial. We will factor it first.

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#math.equation(block: false, alt: "4 n squared plus 4 n plus 1 equals 16")[$#h(0.8em) 4 n^(2) + 4 n + 1 = 16$]]),
  [Factor the perfect square trinomial.], [#math.equation(block: false, alt: "open parenthesis 2 n plus 1 close parenthesis squared equals 16")[$#h(2.25em) attach(( 2 n + 1 ), t: 2) = 16$]],
  [Use the Square Root Property.], [#math.equation(block: false, alt: "2 n plus 1 equals plus or minus the square root of 16")[$#h(3.4em) 2 n + 1 = ± sqrt(16)$]],
  [Simplify the radical.], [#math.equation(block: false, alt: "2 n plus 1 equals plus or minus 4")[$#h(3.4em) 2 n + 1 = ± 4$]],
  [Solve for #math.equation(block: false, alt: "n")[$n$].], [#math.equation(block: false, alt: "2 n equals −1 plus or minus 4")[$#h(5em) 2 n = −1 ± 4$]],
  [Divide each side by 2.], [#math.equation(block: false, alt: "the fraction 2 n over 2 equals the fraction −1 plus or minus 4 over 2")[$#h(5em) frac(2 n, 2) = frac(−1 ± 4, 2)$] #linebreak() 
#math.equation(block: false, alt: "n equals the fraction −1 plus or minus 4 over 2")[$#h(5.6em) n = frac(−1 ± 4, 2)$]],
  [Rewrite to show two solutions.], [#math.equation(block: false, alt: "n equals the fraction −1 plus 4 over 2")[$n = frac(−1 + 4, 2)$], #math.equation(block: false, alt: "n equals the fraction −1 minus 4 over 2")[$n = frac(−1 − 4, 2)$]],
  [Simplify each equation.], [#math.equation(block: false, alt: "n equals the fraction 3 over 2")[$n = frac(3, 2)$], #math.equation(block: false, alt: "n equals minus the fraction 5 over 2")[$#h(2.35em) n = − frac(5, 2)$]],
  [Check: #linebreak()  #linebreak() 
#figure(figph[Two solutions for the equation 4n^2 + 4n + 1 = 16 are verified: n=3/2 and n=-5/2, both satisfying the equation and confirming 16=16.], alt: "Two solutions for the equation 4n^2 + 4n + 1 = 16 are verified: n=3/2 and n=-5/2, both satisfying the equation and confirming 16=16.", caption: none)], [],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve: #math.equation(block: false, alt: "9 m squared minus 12 m plus 4 equals 25 .")[$9 m^(2) − 12 m + 4 = 25 .$]

#solutionbox[
#math.equation(block: true, alt: "m equals the fraction 7 over 3 , m equals −1")[$m = frac(7, 3) , #h(0.5em) m = −1$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve: #math.equation(block: false, alt: "16 n squared plus 40 n plus 25 equals 4 .")[$16 n^(2) + 40 n + 25 = 4 .$]

#solutionbox[
#math.equation(block: true, alt: "n equals minus the fraction 3 over 4 , n equals minus the fraction 7 over 4")[$n = − frac(3, 4) , #h(0.5em) n = − frac(7, 4)$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Access this online resource for additional instruction and practice with using the Square Root Property to solve quadratic equations.

- #link("https://openstax.org/l/37SqRtProp1")[Solving Quadratic Equations: The Square Root Property]
- #link("https://openstax.org/l/37SqRtProp2")[Using the Square Root Property to Solve Quadratic Equations]
]


=== Key Concepts

- Square Root Property
- If #math.equation(block: false, alt: "x squared equals k")[$x^(2) = k$], then #math.equation(block: false, alt: "x equals the square root of k or x equals − the square root of k")[$x = sqrt(k) #h(0.5em) "or" #h(0.5em) x = "−" sqrt(k)$] or #math.equation(block: false, alt: "x equals plus or minus the square root of k")[$x = ± sqrt(k)$]

 #linebreak() 
How to solve a quadratic equation using the square root property.
+ Isolate the quadratic term and make its coefficient one.
+ Use Square Root Property.
+ Simplify the radical.
+ Check the solutions.


==== Practice Makes Perfect

#strong[Solve Quadratic Equations of the Form #emph[ax]#super[2] = #emph[k] Using the Square Root Property]

In the following exercises, solve each equation.

#math.equation(block: true, alt: "a squared equals 49")[$a^(2) = 49$]

#solutionbox[
#math.equation(block: true, alt: "a equals plus or minus 7")[$a = ± 7$]
]

#math.equation(block: true, alt: "b squared equals 144")[$b^(2) = 144$]

#math.equation(block: true, alt: "r squared minus 24 equals 0")[$r^(2) − 24 = 0$]

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

#math.equation(block: true, alt: "t squared minus 75 equals 0")[$t^(2) − 75 = 0$]

#math.equation(block: true, alt: "u squared minus 300 equals 0")[$u^(2) − 300 = 0$]

#solutionbox[
#math.equation(block: true, alt: "u equals plus or minus 10 the square root of 3")[$u = ± 10 sqrt(3)$]
]

#math.equation(block: true, alt: "v squared minus 80 equals 0")[$v^(2) − 80 = 0$]

#math.equation(block: true, alt: "4 m squared equals 36")[$4 m^(2) = 36$]

#solutionbox[
#math.equation(block: true, alt: "m equals plus or minus 3")[$m = ± 3$]
]

#math.equation(block: true, alt: "3 n squared equals 48")[$3 n^(2) = 48$]

#math.equation(block: true, alt: "the fraction 4 over 3 x squared equals 48")[$frac(4, 3) x^(2) = 48$]

#solutionbox[
#math.equation(block: true, alt: "x equals plus or minus 6")[$x = ± 6$]
]

#math.equation(block: true, alt: "the fraction 5 over 3 y squared equals 60")[$frac(5, 3) y^(2) = 60$]

#math.equation(block: true, alt: "x squared plus 25 equals 0")[$x^(2) + 25 = 0$]

#solutionbox[
#math.equation(block: true, alt: "x equals plus or minus 5 i")[$x = ± 5 i$]
]

#math.equation(block: true, alt: "y squared plus 64 equals 0")[$y^(2) + 64 = 0$]

#math.equation(block: true, alt: "x squared plus 63 equals 0")[$x^(2) + 63 = 0$]

#solutionbox[
#math.equation(block: true, alt: "x equals plus or minus 3 the square root of 7 i")[$x = ± 3 sqrt(7) i$]
]

#math.equation(block: true, alt: "y squared plus 45 equals 0")[$y^(2) + 45 = 0$]

#math.equation(block: true, alt: "the fraction 4 over 3 x squared plus 2 equals 110")[$frac(4, 3) x^(2) + 2 = 110$]

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

#math.equation(block: true, alt: "the fraction 2 over 3 y squared minus 8 equals −2")[$frac(2, 3) y^(2) − 8 = −2$]

#math.equation(block: true, alt: "the fraction 2 over 5 a squared plus 3 equals 11")[$frac(2, 5) a^(2) + 3 = 11$]

#solutionbox[
#math.equation(block: true, alt: "a equals plus or minus 2 the square root of 5")[$a = ± 2 sqrt(5)$]
]

#math.equation(block: true, alt: "the fraction 3 over 2 b squared minus 7 equals 41")[$frac(3, 2) b^(2) − 7 = 41$]

#math.equation(block: true, alt: "7 p squared plus 10 equals 26")[$7 p^(2) + 10 = 26$]

#solutionbox[
#math.equation(block: true, alt: "p equals plus or minus the fraction 4 the square root of 7 over 7")[$p = ± frac(4 sqrt(7), 7)$]
]

#math.equation(block: true, alt: "2 q squared plus 5 equals 30")[$2 q^(2) + 5 = 30$]

#math.equation(block: true, alt: "5 y squared minus 7 equals 25")[$5 y^(2) − 7 = 25$]

#solutionbox[
#math.equation(block: true, alt: "y equals plus or minus the fraction 4 the square root of 10 over 5")[$y = ± frac(4 sqrt(10), 5)$]
]

#math.equation(block: true, alt: "3 x squared minus 8 equals 46")[$3 x^(2) − 8 = 46$]

#strong[Solve Quadratic Equations of the Form #emph[a] (#emph[x] − #emph[h])#super[2] = #emph[k] Using the Square Root Property]

In the following exercises, solve each equation.

#math.equation(block: true, alt: "open parenthesis u minus 6 close parenthesis squared equals 64")[$attach(( u − 6 ), t: 2) = 64$]

#solutionbox[
#math.equation(block: true, alt: "u equals 14 , u equals −2")[$u = 14 , u = −2$]
]

#math.equation(block: true, alt: "open parenthesis v plus 10 close parenthesis squared equals 121")[$attach(( v + 10 ), t: 2) = 121$]

#math.equation(block: true, alt: "open parenthesis m minus 6 close parenthesis squared equals 20")[$attach(( m − 6 ), t: 2) = 20$]

#solutionbox[
#math.equation(block: true, alt: "m equals 6 plus or minus 2 the square root of 5")[$m = 6 ± 2 sqrt(5)$]
]

#math.equation(block: true, alt: "open parenthesis n plus 5 close parenthesis squared equals 32")[$attach(( n + 5 ), t: 2) = 32$]

#math.equation(block: true, alt: "open parenthesis r minus the fraction 1 over 2 close parenthesis squared equals the fraction 3 over 4")[$attach(( r − frac(1, 2) ), t: 2) = frac(3, 4)$]

#solutionbox[
#math.equation(block: true, alt: "r equals the fraction 1 over 2 plus or minus the fraction the square root of 3 over 2")[$r = frac(1, 2) ± frac(sqrt(3), 2)$]
]

#math.equation(block: true, alt: "open parenthesis x plus the fraction 1 over 5 close parenthesis squared equals the fraction 7 over 25")[$attach(( x + frac(1, 5) ), t: 2) = frac(7, 25)$]

#math.equation(block: true, alt: "open parenthesis y plus the fraction 2 over 3 close parenthesis squared equals the fraction 8 over 81")[$attach(( y + frac(2, 3) ), t: 2) = frac(8, 81)$]

#solutionbox[
#math.equation(block: true, alt: "y equals minus the fraction 2 over 3 plus or minus the fraction 2 the square root of 2 over 9")[$y = − frac(2, 3) ± frac(2 sqrt(2), 9)$]
]

#math.equation(block: true, alt: "open parenthesis t minus the fraction 5 over 6 close parenthesis squared equals the fraction 11 over 25")[$attach(( t − frac(5, 6) ), t: 2) = frac(11, 25)$]

#math.equation(block: true, alt: "open parenthesis a minus 7 close parenthesis squared plus 5 equals 55")[$attach(( a − 7 ), t: 2) + 5 = 55$]

#solutionbox[
#math.equation(block: true, alt: "a equals 7 plus or minus 5 the square root of 2")[$a = 7 ± 5 sqrt(2)$]
]

#math.equation(block: true, alt: "open parenthesis b minus 1 close parenthesis squared minus 9 equals 39")[$attach(( b − 1 ), t: 2) − 9 = 39$]

#math.equation(block: true, alt: "4 open parenthesis x plus 3 close parenthesis squared minus 5 equals 27")[$4 attach(( x + 3 ), t: 2) − 5 = 27$]

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

#math.equation(block: true, alt: "5 open parenthesis x plus 3 close parenthesis squared minus 7 equals 68")[$5 attach(( x + 3 ), t: 2) − 7 = 68$]

#math.equation(block: true, alt: "open parenthesis 5 c plus 1 close parenthesis squared equals −27")[$attach(( 5 c + 1 ), t: 2) = −27$]

#solutionbox[
#math.equation(block: true, alt: "c equals minus the fraction 1 over 5 plus or minus the fraction 3 the square root of 3 over 5 i")[$c = − frac(1, 5) ± frac(3 sqrt(3), 5) i$]
]

#math.equation(block: true, alt: "open parenthesis 8 d minus 6 close parenthesis squared equals −24")[$attach(( 8 d − 6 ), t: 2) = −24$]

#math.equation(block: true, alt: "open parenthesis 4 x minus 3 close parenthesis squared plus 11 equals −17")[$attach(( 4 x − 3 ), t: 2) + 11 = −17$]

#solutionbox[
#math.equation(block: true, alt: "x equals the fraction 3 over 4 plus or minus the fraction the square root of 7 over 2 i")[$x = frac(3, 4) ± frac(sqrt(7), 2) i$]
]

#math.equation(block: true, alt: "open parenthesis 2 y plus 1 close parenthesis squared minus 5 equals −23")[$attach(( 2 y + 1 ), t: 2) − 5 = −23$]

#math.equation(block: true, alt: "m squared minus 4 m plus 4 equals 8")[$m^(2) − 4 m + 4 = 8$]

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

#math.equation(block: true, alt: "n squared plus 8 n plus 16 equals 27")[$n^(2) + 8 n + 16 = 27$]

#math.equation(block: true, alt: "x squared minus 6 x plus 9 equals 12")[$x^(2) − 6 x + 9 = 12$]

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

#math.equation(block: true, alt: "y squared plus 12 y plus 36 equals 32")[$y^(2) + 12 y + 36 = 32$]

#math.equation(block: true, alt: "25 x squared minus 30 x plus 9 equals 36")[$25 x^(2) − 30 x + 9 = 36$]

#solutionbox[
#math.equation(block: true, alt: "x equals minus the fraction 3 over 5 , x equals the fraction 9 over 5")[$x = − frac(3, 5) , x = frac(9, 5)$]
]

#math.equation(block: true, alt: "9 y squared plus 12 y plus 4 equals 9")[$9 y^(2) + 12 y + 4 = 9$]

#math.equation(block: true, alt: "36 x squared minus 24 x plus 4 equals 81")[$36 x^(2) − 24 x + 4 = 81$]

#solutionbox[
#math.equation(block: true, alt: "x equals minus the fraction 7 over 6 , x equals the fraction 11 over 6")[$x = − frac(7, 6) , x = frac(11, 6)$]
]

#math.equation(block: true, alt: "64 x squared plus 144 x plus 81 equals 25")[$64 x^(2) + 144 x + 81 = 25$]


==== Mixed Practice

In the following exercises, solve using the Square Root Property.

#math.equation(block: true, alt: "2 r squared equals 32")[$2 r^(2) = 32$]

#solutionbox[
#math.equation(block: true, alt: "r equals plus or minus 4")[$r = ± 4$]
]

#math.equation(block: true, alt: "4 t squared equals 16")[$4 t^(2) = 16$]

#math.equation(block: true, alt: "open parenthesis a minus 4 close parenthesis squared equals 28")[$attach(( a − 4 ), t: 2) = 28$]

#solutionbox[
#math.equation(block: true, alt: "a equals 4 plus or minus 2 the square root of 7")[$a = 4 ± 2 sqrt(7)$]
]

#math.equation(block: true, alt: "open parenthesis b plus 7 close parenthesis squared equals 8")[$attach(( b + 7 ), t: 2) = 8$]

#math.equation(block: true, alt: "9 w squared minus 24 w plus 16 equals 1")[$9 w^(2) − 24 w + 16 = 1$]

#solutionbox[
#math.equation(block: true, alt: "w equals 1 , w equals the fraction 5 over 3")[$w = 1 , w = frac(5, 3)$]
]

#math.equation(block: true, alt: "4 z squared plus 4 z plus 1 equals 49")[$4 z^(2) + 4 z + 1 = 49$]

#math.equation(block: true, alt: "a squared minus 18 equals 0")[$a^(2) − 18 = 0$]

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

#math.equation(block: true, alt: "b squared minus 108 equals 0")[$b^(2) − 108 = 0$]

#math.equation(block: true, alt: "open parenthesis p minus the fraction 1 over 3 close parenthesis squared equals the fraction 7 over 9")[$attach(( p − frac(1, 3) ), t: 2) = frac(7, 9)$]

#solutionbox[
#math.equation(block: true, alt: "p equals the fraction 1 over 3 plus or minus the fraction the square root of 7 over 3")[$p = frac(1, 3) ± frac(sqrt(7), 3)$]
]

#math.equation(block: true, alt: "open parenthesis q minus the fraction 3 over 5 close parenthesis squared equals the fraction 3 over 4")[$attach(( q − frac(3, 5) ), t: 2) = frac(3, 4)$]

#math.equation(block: true, alt: "m squared plus 12 equals 0")[$m^(2) + 12 = 0$]

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

#math.equation(block: true, alt: "n squared plus 48 equals 0 .")[$n^(2) + 48 = 0 .$]

#math.equation(block: true, alt: "u squared minus 14 u plus 49 equals 72")[$u^(2) − 14 u + 49 = 72$]

#solutionbox[
#math.equation(block: true, alt: "u equals 7 plus or minus 6 the square root of 2")[$u = 7 ± 6 sqrt(2)$]
]

#math.equation(block: true, alt: "v squared plus 18 v plus 81 equals 50")[$v^(2) + 18 v + 81 = 50$]

#math.equation(block: true, alt: "open parenthesis m minus 4 close parenthesis squared plus 3 equals 15")[$attach(( m − 4 ), t: 2) + 3 = 15$]

#solutionbox[
#math.equation(block: true, alt: "m equals 4 plus or minus 2 the square root of 3")[$m = 4 ± 2 sqrt(3)$]
]

#math.equation(block: true, alt: "open parenthesis n minus 7 close parenthesis squared minus 8 equals 64")[$attach(( n − 7 ), t: 2) − 8 = 64$]

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

#solutionbox[
#math.equation(block: true, alt: "x equals −3 , x equals −7")[$x = −3 , x = −7$]
]

#math.equation(block: true, alt: "open parenthesis y minus 4 close parenthesis squared equals 64")[$attach(( y − 4 ), t: 2) = 64$]

#math.equation(block: true, alt: "6 c squared plus 4 equals 29")[$6 c^(2) + 4 = 29$]

#solutionbox[
#math.equation(block: true, alt: "c equals plus or minus the fraction 5 the square root of 6 over 6")[$c = ± frac(5 sqrt(6), 6)$]
]

#math.equation(block: true, alt: "2 d squared minus 4 equals 77")[$2 d^(2) − 4 = 77$]

#math.equation(block: true, alt: "open parenthesis x minus 6 close parenthesis squared plus 7 equals 3")[$attach(( x − 6 ), t: 2) + 7 = 3$]

#solutionbox[
#math.equation(block: true, alt: "x equals 6 plus or minus 2 i")[$x = 6 ± 2 i$]
]

#math.equation(block: true, alt: "open parenthesis y minus 4 close parenthesis squared plus 10 equals 9")[$attach(( y − 4 ), t: 2) + 10 = 9$]


==== Writing Exercises

In your own words, explain the Square Root Property.

#solutionbox[
Answers will vary.
]

In your own words, explain how to use the Square Root Property to solve the quadratic equation #math.equation(block: false, alt: "open parenthesis x plus 2 close parenthesis squared equals 16")[$attach(( x + 2 ), t: 2) = 16$].


==== Self Check

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

#figure(figph[Table checklist for evaluating understanding. It asks students to rate their ability to solve two types of quadratic equations using the Square Root Property: first, a times x squared equals k; second, a times the square of x minus h equals k. Response options are "Confidently," "with some help," or "No, I don’t get it."], alt: "Table checklist for evaluating understanding. It asks students to rate their ability to solve two types of quadratic equations using the Square Root Property: first, a times x squared equals k; second, a times the square of x minus h equals k. Response options are \"Confidently,\" \"with some help,\" or \"No, I don’t get it.\"", caption: none)

Choose how would you respond to the statement “I can solve quadratic equations of the form a times the square of x minus h equals k using the Square Root Property.” “Confidently,” “with some help,” or “No, I don’t get it.”

ⓑ If most of your checks were:

#strong[…confidently.] Congratulations! You have achieved the objectives in this section. Reflect on the study skills you used so that you can continue to use them. What did you do to become confident of your ability to do these things? Be specific.

#strong[…with some help.] This must be addressed quickly because topics you do not master become potholes in your road to success. In math every topic builds upon previous work. It is important to make sure you have a strong foundation before you move on. Whom can you ask for help?Your fellow classmates and instructor are good resources. Is there a place on campus where math tutors are available? Can your study skills be improved?

#strong[…no - I don’t get it!] This is a warning sign and you must not ignore it. You should get help right away or you will quickly be overwhelmed. See your instructor as soon as you can to discuss your situation. Together you can come up with a plan to get you the help you need.