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== 10.1#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 75")[$sqrt(75)$]. #linebreak() 
If you missed this problem, review .

#solutionbox[
#math.equation(block: true, alt: "5 the square root of 3")[$5 sqrt(3)$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Simplify: #math.equation(block: false, alt: "the square root of the fraction 64 over 3")[$sqrt(frac(64, 3))$]. #linebreak() 
If you missed this problem, review .

#solutionbox[
#math.equation(block: true, alt: "the fraction 8 the square root of 3 over 3")[$frac(8 sqrt(3), 3)$]
]
]

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

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

Quadratic equations are equations of the form #math.equation(block: false, alt: "a x squared plus b x plus c equals 0")[$a x^(2) + b x + c = 0$], where #math.equation(block: false, alt: "a not equal to 0")[$a ≠ 0$]. They differ from linear equations by including a term with the variable raised to the second power. We use different methods to solve #strong[quadratic equation]s 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 use three other methods to solve quadratic equations.


=== Solve Quadratic Equations of the Form #emph[ax]#super[2] = #emph[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 #math.equation(block: false, alt: "x squared equals 9")[$x^(2) = 9$].

#math.equation(block: true, alt: "x squared, equals, 9; Put the equation in standard form., x squared minus 9, equals, 0; Factor the left side., open parenthesis x minus 3 close parenthesis open parenthesis x plus 3 close parenthesis, equals, 0; Use the Zero Product Property., open parenthesis x minus 3 close parenthesis equals 0 , open parenthesis x plus 3 close parenthesis, equals, 0; Solve each equation., x equals 3 , x, equals, −3; Combine the two solutions into plus or minus form., x, equals, plus or minus 3; (The solution is read ‘ x is equal to positive or negative 3.’)")[$&  &  & x^(2) & = & 9 \ "Put the equation in standard form." &  &  & x^(2) − 9 & = & 0 \ "Factor the left side." &  &  & ( x − 3 ) ( x + 3 ) & = & 0 \ "Use the Zero Product Property." &  &  & ( x − 3 ) = 0 , #h(0.5em) ( x + 3 ) & = & 0 \ "Solve each equation." &  &  & x = 3 , #h(2.8em) x & = & −3 \ "Combine the two solutions into" #h(0.2em) ± #h(0.2em) "form." &  &  & x & = & ± #h(0.2em) 3 \ "(The solution is read" #h(0.2em) ‘ x #h(0.2em) "is equal to positive or negative 3.’)" &  &$]We can easily use factoring to find the solutions of similar equations, like #math.equation(block: false, alt: "x squared equals 16")[$x^(2) = 16$] and #math.equation(block: false, alt: "x squared equals 25")[$x^(2) = 25$], because 16 and 25 are perfect squares. But what happens when we have an equation like #math.equation(block: false, alt: "x squared equals 7")[$x^(2) = 7$]? Since 7 is not a perfect square, we cannot solve the equation by factoring.

These equations are all of the form #math.equation(block: false, alt: "x squared equals k")[$x^(2) = k$]. #linebreak() We 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 .")[$"If" #h(0.2em) n^(2) = m , #h(0.2em) "then" #h(0.2em) n #h(0.2em) "is a square root of" #h(0.2em) m .$]This leads to the #strong[Square Root Property].

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

If #math.equation(block: false, alt: "x squared equals k")[$x^(2) = k$], and #math.equation(block: false, alt: "k greater than or equal to 0")[$k ≥ 0$], then #math.equation(block: false, alt: "x equals the square root of k or x equals − the square root of k")[$x = sqrt(k) #h(0.2em) "or" #h(0.2em) x = "−" sqrt(k)$].
]

Notice that the #strong[Square Root Property] gives two solutions to an equation of the form #math.equation(block: false, alt: "x squared equals k")[$x^(2) = k$]: 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")[$x = ± #h(0.2em) sqrt(k)$].

Now, we will solve the equation #math.equation(block: false, alt: "x squared equals 9")[$x^(2) = 9$] again, this time using the Square Root Property.

#math.equation(block: true, alt: "x squared, equals, 9; Use the Square Root Property., x, equals, plus or minus the square root of 9; Simplify the radical., x, equals, plus or minus 3; Rewrite to show the two solutions., x equals 3 , x, equals, −3")[$&  &  & x^(2) & = & 9 \ "Use the Square Root Property." &  &  & x & = & ± #h(0.2em) sqrt(9) \ "Simplify the radical." &  &  & x & = & ± #h(0.2em) 3 \ "Rewrite to show the two solutions." &  &  & x = 3 , x & = & −3$]What happens when the constant is not a perfect square? Let’s use the Square Root Property to solve the equation #math.equation(block: false, alt: "x squared equals 7")[$x^(2) = 7$].

#math.equation(block: true, alt: "Use the Square Root Property., x squared, equals, 7; x, equals, plus or minus the square root of 7; Rewrite to show two solutions., x equals the square root of 7 , x equals − the square root of 7; We cannot simplify the square root of 7 , so we leave the answer as a radical.")[$"Use the Square Root Property." &  &  & x^(2) & = & 7 \ x & = & ± #h(0.2em) sqrt(7) \ "Rewrite to show two solutions." &  &  & x = sqrt(7) , #h(1em) x = "−" sqrt(7) \ "We cannot simplify" #h(0.2em) sqrt(7) , #h(0.2em) "so we leave the answer as a radical." &  &$]#examplebox("Example 1")[][
Solve: #math.equation(block: false, alt: "x squared equals 169")[$x^(2) = 169$].

#solutionbox[
#math.equation(block: true, alt: "Use the Square Root Property.; Simplify the radical., x squared, equals, 169; x, equals, plus or minus the square root of 169; x, equals, plus or minus 13; Rewrite to show two solutions., x equals 13 , x equals −13")[$"Use the Square Root Property." \ "Simplify the radical." &  &  & #h(4em) x^(2) & = & 169 \ x & = & ± #h(0.2em) sqrt(169) \ x & = & ± #h(0.2em) 13 \ "Rewrite to show two solutions." &  &  & #h(4em) x = 13 , #h(1em) x = −13$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve: #math.equation(block: false, alt: "x squared equals 81")[$x^(2) = 81$].

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

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve: #math.equation(block: false, alt: "y squared equals 121")[$y^(2) = 121$].

#solutionbox[
#math.equation(block: true, alt: "y equals 11 , y equals −11")[$y = 11 , y = −11$]
]
]

#examplebox("Example 2")[How to Solve a Quadratic Equation of the Form #math.equation(block: false, alt: "a x squared equals k")[$attach(a x, t: 2) = k$] Using the Square Root Property][
Solve: #math.equation(block: false, alt: "x squared minus 48 equals 0")[$x^(2) − 48 = 0$].

#solutionbox[
#figure(figph[The image shows the given equation, x squared minus 48 equals zero. Step one is to isolate the quadratic term and make its coefficient one so add 48 to both sides of the equation to get x squared by itself.], alt: "The image shows the given equation, x squared minus 48 equals zero. Step one is to isolate the quadratic term and make its coefficient one so add 48 to both sides of the equation to get x squared by itself.", caption: none)

#figure(figph[Step two is to use the Square Root Property to get x equals plus or minus the square root of 48.], alt: "Step two is to use the Square Root Property to get x equals plus or minus the square root of 48.", caption: none)

#figure(figph[Step three, simplify the square root of 48 by writing 48 as the product of 16 and three. The square root of 16 is four. The simplified solution is x equals plus or minus four square root of three.], alt: "Step three, simplify the square root of 48 by writing 48 as the product of 16 and three. The square root of 16 is four. The simplified solution is x equals plus or minus four square root of three.", caption: none)

#figure(figph[Step four, check the solutions by substituting each solution into the original equation. When x equals four square root of three, replace x in the original equation with four square root of three to get four square root of three squared minus 48 equals zero. Simplify the left side to get 16 times three minus 48 equals zero which simplifies further to zero equals zero, a true statement. When x equals negative four square root of three, replace x in the original equation with negative four square root of three to get negative four square root of three squared minus 48 equals zero. Simplify the left side to get 16 times three minus 48 equals zero which simplifies further to zero equals zero, also a true statement.], alt: "Step four, check the solutions by substituting each solution into the original equation. When x equals four square root of three, replace x in the original equation with four square root of three to get four square root of three squared minus 48 equals zero. Simplify the left side to get 16 times three minus 48 equals zero which simplifies further to zero equals zero, a true statement. When x equals negative four square root of three, replace x in the original equation with negative four square root of three to get negative four square root of three squared minus 48 equals zero. Simplify the left side to get 16 times three minus 48 equals zero which simplifies further to zero equals zero, also a true statement.", caption: none)
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve: #math.equation(block: false, alt: "x squared minus 50 equals 0")[$x^(2) − 50 = 0$].

#solutionbox[
#math.equation(block: true, alt: "x equals 5 the square root of 2 , x equals −5 the square root of 2")[$x = 5 sqrt(2) , x = −5 sqrt(2)$]
]
]

#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")[$y = 3 sqrt(3) , y = −3 sqrt(3)$]
]
]

#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.
]

To use the Square Root Property, the coefficient of the variable term must equal 1. In the next example, we must divide both sides of the equation by 5 before using the Square Root Property.

#examplebox("Example 3")[][
Solve: #math.equation(block: false, alt: "5 m squared equals 80")[$5 m^(2) = 80$].

#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([The quadratic term is isolated.], [#math.equation(block: false, alt: "5 m squared equals 80")[$#h(0.1em) 5 m^(2) = 80$]]),
  [Divide by 5 to make its cofficient 1.], [#math.equation(block: false, alt: "the fraction 5 m squared over 5 equals the fraction 80 over 5")[$frac(5 m^(2), 5) = frac(80, 5)$]],
  [Simplify.], [#math.equation(block: false, alt: "m squared equals 16")[$#h(0.6em) m^(2) = 16$]],
  [Use the Square Root Property.], [#math.equation(block: false, alt: "m equals plus or minus the square root of 16")[$#h(1.1em) m = ± #h(0.2em) sqrt(16)$]],
  [Simplify the radical.], [#math.equation(block: false, alt: "m equals plus or minus 4")[$#h(1.1em) m = ± #h(0.2em) 4$]],
  [Rewrite to show two solutions.], [#math.equation(block: false, alt: "m equals 4 , m equals − 4")[$m = 4 , m = "−" 4$]],
  [Check the solutions. #linebreak() #figure(figph[Verifying that both m=4 and m=-4 are solutions to the equation 5m^2=80, as both positive and negative values yield 80=80 when squared and multiplied by 5.], alt: "Verifying that both m=4 and m=-4 are solutions to the equation 5m^2=80, as both positive and negative values yield 80=80 when squared and multiplied by 5.", 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: "3 z squared equals 108")[$3 z^(2) = 108$].

#solutionbox[
#math.equation(block: true, alt: "z equals 6 , z equals −6")[$z = 6 , z = −6$]
]
]

The Square Root Property started by stating, ‘If #math.equation(block: false, alt: "x squared equals k")[$x^(2) = k$], and #math.equation(block: false, alt: "k greater than or equal to 0")[$k ≥ 0$]’. 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 4")[][
Solve: #math.equation(block: false, alt: "q squared plus 24 equals 0")[$q^(2) + 24 = 0$].

#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#math.equation(block: false, alt: "q squared plus 24 equals 0")[$q^(2) + 24 = 0$]]),
  [Isolate the quadratic term.], [#math.equation(block: false, alt: "q squared equals −24")[$#h(2em) q^(2) = −24$]],
  [Use the Square Root Property.], [#math.equation(block: false, alt: "q equals plus or minus the square root of −24")[$#h(2.5em) q = ± #h(0.2em) sqrt(−24)$]],
  [The #math.equation(block: false, alt: "the square root of −24")[$sqrt(−24)$] is not a real number.], [There is no real solution.],
))
]
]

#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[
no real solution
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve: #math.equation(block: false, alt: "d squared plus 81 equals 0")[$d^(2) + 81 = 0$].

#solutionbox[
no real solution
]
]

Remember, we first isolate the quadratic term and then make the coefficient equal to one.

#examplebox("Example 5")[][
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.], [#math.equation(block: false, alt: "the fraction 2 over 3 u squared equals 12")[$#h(1.7em) frac(2, 3) u^(2) = 12$]],
  [Multiply by #math.equation(block: false, alt: "the fraction 3 over 2")[$frac(3, 2)$] to make the coefficient 1.], [#math.equation(block: false, alt: "the fraction 3 over 2 times the fraction 2 over 3 u squared equals the fraction 3 over 2 times 12")[$#h(0.5em) frac(3, 2) · frac(2, 3) u^(2) = frac(3, 2) · 12$]],
  [Simplify.], [#math.equation(block: false, alt: "u squared equals 18")[$#h(2.2em) u^(2) = 18$]],
  [Use the Square Root Property.], [#math.equation(block: false, alt: "u equals plus or minus the square root of 18")[$#h(2.7em) u = ± #h(0.2em) sqrt(18)$]],
  [Simplify the radical.], [#math.equation(block: false, alt: "u equals plus or minus the square root of 9 the square root of 2")[$#h(2.7em) u = ± #h(0.2em) sqrt(9) sqrt(2)$]],
  [Simplify.], [#math.equation(block: false, alt: "u equals plus or minus 3 the square root of 2")[$#h(2.7em) u = ± #h(0.2em) 3 sqrt(2)$]],
  [Rewrite to show two solutions.], [#math.equation(block: false, alt: "u equals 3 the square root of 2 , u equals − 3 the square root of 2")[$u = 3 sqrt(2) , u = "−" 3 sqrt(2)$]],
  [Check. #linebreak() #figure(figph[Two step-by-step mathematical calculations demonstrating the verification of both positive and negative solutions for the quadratic equation (2/3)u^2 + 5 = 17, confirming 17 = 17 in both cases.], alt: "Two step-by-step mathematical calculations demonstrating the verification of both positive and negative solutions for the quadratic equation (2/3)u^2 + 5 = 17, confirming 17 = 17 in both cases.", 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")[$x = 2 sqrt(10) , x = −2 sqrt(10)$]
]
]

#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")[$y = 2 sqrt(7) , y = −2 sqrt(7)$]
]
]

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

#examplebox("Example 6")[][
Solve: #math.equation(block: false, alt: "2 c squared minus 4 equals 45")[$2 c^(2) − 4 = 45$].

#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([#linebreak()  #linebreak()  #linebreak() Isolate the quadratic term. #linebreak()  #linebreak() Divide by 2 to make the coefficient 1. #linebreak()  #linebreak()  #linebreak()  #linebreak() Simplify. #linebreak()  #linebreak() Use the Square Root Property. #linebreak()  #linebreak() Simplify the radical. #linebreak()  #linebreak() Rationalize the denominator. #linebreak()  #linebreak() Simplify.], [#math.equation(block: false, alt: "2 c squared minus 4, equals, 45; 2 c squared, equals, 49; the fraction 2 c squared over 2, equals, the fraction 49 over 2; c squared, equals, the fraction 49 over 2; c, equals, plus or minus the square root of the fraction 49 over 2; c, equals, plus or minus the fraction the square root of 49 over the square root of 2; c, equals, plus or minus the fraction the square root of 49 times the square root of 2 over the square root of 2 times the square root of 2; c, equals, plus or minus the fraction 7 the square root of 2 over 2")[$#h(4em) 2 c^(2) − 4 & = & 45 \ #h(4em) 2 c^(2) & = & 49 \ #h(4em) frac(2 c^(2), 2) & = & frac(49, 2) \ #h(4em) c^(2) & = & frac(49, 2) \ #h(4em) c & = & ± #h(0.2em) sqrt(frac(49, 2)) \ #h(4em) c & = & ± #h(0.2em) frac(sqrt(49), sqrt(2)) \ #h(4em) c & = & ± #h(0.2em) frac(sqrt(49) · sqrt(2), sqrt(2) · sqrt(2)) \ #h(4em) c & = & ± #h(0.2em) frac(7 sqrt(2), 2)$]]),
  [Rewrite to show two solutions.], [#math.equation(block: false, alt: "c equals the fraction 7 the square root of 2 over 2 , c equals minus the fraction 7 the square root of 2 over 2")[$c = frac(7 sqrt(2), 2) , #h(1em) c = − frac(7 sqrt(2), 2)$]],
  [Check. 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")[$r = frac(6 sqrt(5), 5) , r = − frac(6 sqrt(5), 5)$]
]
]

#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")[$t = frac(8 sqrt(3), 3) , t = − frac(8 sqrt(3), 3)$]
]
]


=== 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 Square Root Property to solve an equation like #math.equation(block: false, alt: "open parenthesis x minus 3 close parenthesis squared equals 16")[$attach(( x − 3 ), t: 2) = 16$], too. We will treat the whole binomial, #math.equation(block: false, alt: "open parenthesis x minus 3 close parenthesis")[$( x − 3 )$], as the quadratic term.

#examplebox("Example 7")[][
Solve: #math.equation(block: false, alt: "open parenthesis x minus 3 close parenthesis squared equals 16")[$attach(( x − 3 ), t: 2) = 16$].

#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#math.equation(block: false, alt: "open parenthesis x minus 3 close parenthesis squared equals 16")[$attach(( x − 3 ), t: 2) = 16$]]),
  [Use the Square Root Property.], [#math.equation(block: false, alt: "x minus 3 equals plus or minus the square root of 16")[$#h(1.1em) x − 3 = ± #h(0.2em) sqrt(16)$]],
  [Simplify.], [#math.equation(block: false, alt: "x minus 3 equals plus or minus 4")[$#h(1.1em) x − 3 = ± #h(0.2em) 4$]],
  [Write as two equations.], [#math.equation(block: false, alt: "x minus 3 equals 4 , x minus 3 equals − 4")[$x − 3 = 4 , x − 3 = "−" 4$]],
  [Solve.], [#math.equation(block: false, alt: "x equals 7 , x equals − 1")[$#h(1.6em) x = 7 , x = "−" 1$]],
  [Check. #linebreak() #figure(figph[Two math problems are solved: (7-3)^2=16 and (-1-3)^2=16. Both simplify to (4)^2=16 and (-4)^2=16 respectively, ultimately confirming 16=16, illustrating squaring positive and negative numbers.], alt: "Two math problems are solved: (7-3)^2=16 and (-1-3)^2=16. Both simplify to (4)^2=16 and (-4)^2=16 respectively, ultimately confirming 16=16, illustrating squaring positive and negative numbers.", caption: none)], [],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve: #math.equation(block: false, alt: "open parenthesis q plus 5 close parenthesis squared equals 1")[$attach(( q + 5 ), t: 2) = 1$].

#solutionbox[
#math.equation(block: true, alt: "q equals −6 , q equals −4")[$q = −6 , q = −4$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve: #math.equation(block: false, alt: "open parenthesis r minus 3 close parenthesis squared equals 25")[$attach(( r − 3 ), t: 2) = 25$].

#solutionbox[
#math.equation(block: true, alt: "r equals 8 , r equals −2")[$r = 8 , r = −2$]
]
]

#examplebox("Example 8")[][
Solve: #math.equation(block: false, alt: "open parenthesis y minus 7 close parenthesis squared equals 12")[$attach(( y − 7 ), t: 2) = 12$].

#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#math.equation(block: false, alt: "open parenthesis y minus 7 close parenthesis squared equals 12")[$attach(( y − 7 ), t: 2) = 12$]]),
  [Use the Square Root Property.], [#math.equation(block: false, alt: "y minus 7 equals plus or minus the square root of 12")[$#h(0.9em) y − 7 = ± #h(0.2em) sqrt(12)$]],
  [Simplify the radical.], [#math.equation(block: false, alt: "y minus 7 equals plus or minus 2 the square root of 3")[$#h(0.9em) y − 7 = ± #h(0.2em) 2 sqrt(3)$]],
  [Solve for #emph[y].], [#math.equation(block: false, alt: "y equals 7 plus or minus 2 the square root of 3")[$#h(2.5em) y = 7 ± 2 sqrt(3)$]],
  [Rewrite to show two solutions.], [#math.equation(block: false, alt: "y equals 7 plus 2 the square root of 3 , y equals 7 minus 2 the square root of 3")[$y = 7 + 2 sqrt(3) , y = 7 − 2 sqrt(3)$]],
  [Check. #linebreak() #figure(figph[Two columns of mathematical steps verify the solutions for the equation (y-7)^2 = 12. Both y = 7 + 2sqrt(3) and y = 7 - 2sqrt(3) are shown to correctly satisfy the equation, resulting in 12 = 12.], alt: "Two columns of mathematical steps verify the solutions for the equation (y-7)^2 = 12. Both y = 7 + 2sqrt(3) and y = 7 - 2sqrt(3) are shown to correctly satisfy the equation, resulting in 12 = 12.", caption: none)], [],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve: #math.equation(block: false, alt: "open parenthesis a minus 3 close parenthesis squared equals 18")[$attach(( a − 3 ), t: 2) = 18$].

#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")[$a = 3 + 3 sqrt(2) , a = 3 − 3 sqrt(2)$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve: #math.equation(block: false, alt: "open parenthesis b plus 2 close parenthesis squared equals 40")[$attach(( b + 2 ), t: 2) = 40$].

#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")[$b = −2 + 2 sqrt(10) , b = −2 − 2 sqrt(10)$]
]
]

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

#examplebox("Example 9")[][
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[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([#linebreak()  #linebreak()  #linebreak()  #linebreak() Use the Square Root Property. #linebreak()  #linebreak() Rewrite the radical as a fraction of square roots. #linebreak()  #linebreak()  #linebreak() Simplify the radical. #linebreak()  #linebreak()  #linebreak() Solve for #emph[x.]], [#math.equation(block: false, alt: "open parenthesis x minus the fraction 1 over 2 close parenthesis squared, equals, the fraction 5 over 4; x minus the fraction 1 over 2, equals, plus or minus the square root of the fraction 5 over 4; x minus the fraction 1 over 2, equals, plus or minus the fraction the square root of 5 over the square root of 4; x minus the fraction 1 over 2, equals, plus or minus the fraction the square root of 5 over 2; x, equals, the fraction 1 over 2 plus or minus the fraction the square root of 5 over 2")[$attach(( x − frac(1, 2) ), t: 2) & = & frac(5, 4) \ x − frac(1, 2) & = & ± #h(0.2em) sqrt(frac(5, 4)) \ x − frac(1, 2) & = & ± #h(0.2em) frac(sqrt(5), sqrt(4)) \ x − frac(1, 2) & = & ± #h(0.2em) frac(sqrt(5), 2) \ x & = & frac(1, 2) ± frac(sqrt(5), 2)$]]),
  [Rewrite to show two solutions.], [#math.equation(block: false, alt: "x equals the fraction 1 over 2 plus the fraction the square root of 5 over 2 , x equals the fraction 1 over 2 minus the fraction the square root of 5 over 2")[$x = frac(1, 2) + frac(sqrt(5), 2) , #h(1em) x = frac(1, 2) − frac(sqrt(5), 2)$]],
  [Check. 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 3 close parenthesis squared equals the fraction 5 over 9 .")[$attach(( x − frac(1, 3) ), t: 2) = frac(5, 9) .$]

#solutionbox[
#math.equation(block: true, 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")[$x = frac(1, 3) + frac(sqrt(5), 3) , x = frac(1, 3) − frac(sqrt(5), 3)$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve: #math.equation(block: false, alt: "open parenthesis y minus 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 the fraction 3 over 4 plus the fraction the square root of 7 over 4 , y equals the fraction 3 over 4 minus the fraction the square root of 7 over 4")[$y = frac(3, 4) + frac(sqrt(7), 4) , y = frac(3, 4) − frac(sqrt(7), 4)$]
]
]

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

#examplebox("Example 10")[][
Solve: #math.equation(block: false, alt: "open parenthesis x minus 2 close parenthesis squared plus 3 equals 30")[$attach(( x − 2 ), t: 2) + 3 = 30$].

#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#math.equation(block: false, alt: "open parenthesis x minus 2 close parenthesis squared plus 3, equals, 30")[$#h(1.3em) attach(( x − 2 ), t: 2) + 3 & = & 30$]]),
  [Isolate the binomial term.], [#math.equation(block: false, alt: "open parenthesis x minus 2 close parenthesis squared, equals, 27")[$#h(3em) attach(( x − 2 ), t: 2) & = & 27$]],
  [Use the Square Root Property.], [#math.equation(block: false, alt: "x minus 2, equals, plus or minus the square root of 27")[$#h(4.3em) x − 2 & = & ± #h(0.2em) sqrt(27)$]],
  [Simplify the radical.], [#math.equation(block: false, alt: "x minus 2, equals, plus or minus 3 the square root of 3")[$#h(4.3em) x − 2 & = & ± #h(0.2em) 3 sqrt(3)$]],
  [Solve for #emph[x].], [#math.equation(block: false, alt: "x, equals, 2 plus or minus 3 the square root of 3")[$#h(6em) x & = & 2 ± 3 sqrt(3)$]],
  [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")[$#h(2em) x = 2 + 3 sqrt(3) , #h(1em) x = 2 − 3 sqrt(3)$]],
  [Check. We leave the check for you.], [],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve: #math.equation(block: false, alt: "open parenthesis a minus 5 close parenthesis squared plus 4 equals 24")[$attach(( a − 5 ), t: 2) + 4 = 24$].

#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")[$a = 5 + 2 sqrt(5) , a = 5 − 2 sqrt(5)$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve: #math.equation(block: false, alt: "open parenthesis b minus 3 close parenthesis squared minus 8 equals 24")[$attach(( b − 3 ), t: 2) − 8 = 24$].

#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")[$b = 3 + 4 sqrt(2) , b = 3 − 4 sqrt(2)$]
]
]

#examplebox("Example 11")[][
Solve: #math.equation(block: false, alt: "open parenthesis 3 v minus 7 close parenthesis squared equals −12")[$attach(( 3 v − 7 ), t: 2) = −12$].

#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([#linebreak()  #linebreak()  #linebreak() Use the Square Root Property.], [#math.equation(block: false, alt: "open parenthesis 3 v minus 7 close parenthesis squared, equals, −12; 3 v minus 7, equals, plus or minus the square root of −12")[$attach(( 3 v − 7 ), t: 2) & = & −12 \ 3 v − 7 & = & ± #h(0.2em) sqrt(−12)$]]),
  [The #math.equation(block: false, alt: "the square root of −12")[$sqrt(−12)$] is not a real number.], [There is no real solution.],
))
]
]

#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[
no real solution
]
]

#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[
no real solution
]
]

The left sides of the equations in the next two examples do not seem to be of the form #math.equation(block: false, alt: "a open parenthesis x minus h close parenthesis squared")[$a attach(( x − h ), t: 2)$]. But they are perfect square trinomials, so we will factor to put them in the form we need.

#examplebox("Example 12")[][
Solve: #math.equation(block: false, alt: "p squared minus 10 p plus 25 equals 18")[$p^(2) − 10 p + 25 = 18$].

#solutionbox[
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([#linebreak()  #linebreak()  #linebreak() Factor the perfect square trinomial. #linebreak()  #linebreak() Use the Square Root Property. #linebreak()  #linebreak() Simplify the radical. #linebreak()  #linebreak() Solve for #emph[p].], [#math.equation(block: false, alt: "p squared minus 10 p plus 25, equals, 18; open parenthesis p minus 5 close parenthesis squared, equals, 18; p minus 5, equals, plus or minus the square root of 18; p minus 5, equals, plus or minus 3 the square root of 2; p, equals, 5 plus or minus 3 the square root of 2")[$p^(2) − 10 p + 25 & = & 18 \  \ attach(( p − 5 ), t: 2) & = & 18 \  \ p − 5 & = & ± #h(0.2em) sqrt(18) \  \ p − 5 & = & ± #h(0.2em) 3 sqrt(2) \  \ p & = & 5 ± 3 sqrt(2)$]]),
  [Rewrite to show two solutions.], [#math.equation(block: false, alt: "p equals 5 plus 3 the square root of 2 , p equals 5 minus 3 the square root of 2")[$p = 5 + 3 sqrt(2 ",") #h(1em) p = 5 − 3 sqrt(2)$]],
  [Check. We leave the check for you.], [],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve: #math.equation(block: false, 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 2 the square root of 3 , x equals 3 minus 2 the square root of 3")[$x = 3 + 2 sqrt(3) , x = 3 − 2 sqrt(3)$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve: #math.equation(block: false, alt: "y squared plus 12 y plus 36 equals 32")[$y^(2) + 12 y + 36 = 32$].

#solutionbox[
#math.equation(block: true, alt: "y equals −6 plus 4 the square root of 2 , y equals −6 minus 4 the square root of 2")[$y = −6 + 4 sqrt(2) , y = −6 − 4 sqrt(2)$]
]
]

#examplebox("Example 13")[][
Solve: #math.equation(block: false, alt: "4 n squared plus 4 n plus 1 equals 16")[$4 n^(2) + 4 n + 1 = 16$].

#solutionbox[
Again, 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")[$4 n^(2) + 4 n + 1 = 16$]]),
  [Factor the perfect square trinomial.], [#math.equation(block: false, alt: "open parenthesis 2 n plus 1 close parenthesis squared equals 16")[$#h(1.5em) attach(( 2 n + 1 ), t: 2) = 16$]],
  [Use the Square Root Property.], [#math.equation(block: false, alt: "2 n plus 1 equals plus or minus the square root of 16")[$#h(2.6em) 2 n + 1 = ± #h(0.2em) sqrt(16)$]],
  [Simplify the radical.], [#math.equation(block: false, alt: "2 n plus 1 equals plus or minus 4")[$#h(2.6em) 2 n + 1 = ± #h(0.2em) 4$]],
  [Solve for #emph[n].], [#math.equation(block: false, alt: "2 n equals − 1 plus or minus 4")[$#h(4.2em) 2 n = "−" 1 ± 4$]],
  [Divide each side by 2.], [#math.equation(block: false, alt: "the fraction 2 n over 2, equals, the fraction minus 1 plus or minus 4 over 2; n, equals, the fraction minus 1 plus or minus 4 over 2")[$#h(3.6em) frac(2 n, 2) & = & frac(− 1 ± 4, 2) \ n & = & frac(− 1 ± 4, 2)$]],
  [Rewrite to show two solutions.], [#math.equation(block: false, alt: "n equals the fraction minus 1 plus 4 over 2 , n equals the fraction minus 1 minus 4 over 2")[$n = frac(− 1 + 4, 2) , n = frac(− 1 − 4, 2)$]],
  [Simplify each equation.], [#math.equation(block: false, alt: "n equals the fraction 3 over 2 , n equals minus the fraction 5 over 2")[$#h(2.4em) n = frac(3, 2) , n = − frac(5, 2)$]],
  [Check. #linebreak() #figure(figph[Verification of two solutions (n=3/2 and n=-5/2) for the quadratic equation 4n^2 + 4n + 1 = 16. Both substitutions lead to the correct equality 16 = 16.], alt: "Verification of two solutions (n=3/2 and n=-5/2) for the quadratic equation 4n^2 + 4n + 1 = 16. Both substitutions lead to the correct equality 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 minus 1 , m equals the fraction 7 over 3")[$m = − 1 , m = frac(7, 3)$]
]
]

#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")[$n = − frac(3, 4) , n = − frac(7, 4)$]
]
]

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

- #link("https://openstax.org/l/25Solvebysqroot")[Solving Quadratic Equations: Solving by Taking Square Roots]
- #link("https://openstax.org/l/25Usesqroots")[Using Square Roots to Solve Quadratic Equations]
- #link("https://openstax.org/l/25Sqrtproperty")[Solving Quadratic Equations: The Square Root Method]
]


=== Key Concepts

- Square Root Property #linebreak() If #math.equation(block: false, alt: "x squared equals k")[$x^(2) = k$], and #math.equation(block: false, alt: "k greater than or equal to 0")[$k ≥ 0$], then #math.equation(block: false, alt: "x equals the square root of k or x equals − the square root of k")[$x = sqrt(k) #h(0.2em) "or" #h(0.2em) x = "−" sqrt(k)$].


==== Practice Makes Perfect

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

In the following exercises, solve the following quadratic equations.

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

#solutionbox[
#math.equation(block: true, alt: "a equals plus or minus 7")[$a = ± #h(0.2em) 7$]
]

#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")[$r = ± #h(0.2em) 2 sqrt(6)$]
]

#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")[$u = ± #h(0.2em) 10 sqrt(3)$]
]

#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")[$m = ± #h(0.2em) 3$]
]

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

#math.equation(block: true, alt: "x squared plus 20 equals 0")[$x^(2) + 20 = 0$]

#solutionbox[
no real solution
]

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

#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")[$a = ± #h(0.2em) 2 sqrt(5)$]
]

#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")[$p = ± #h(0.2em) frac(4 sqrt(7), 7)$]
]

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

#strong[Solve Quadratic Equations of the Form] #math.equation(block: false, alt: "a open parenthesis x minus h close parenthesis squared equals k")[$a attach(( x − h ), t: 2) = k$] #strong[Using the Square Root Property]

In the following exercises, solve the following quadratic equations.

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

#solutionbox[
#math.equation(block: true, alt: "x equals 1 , x equals −5")[$x = 1 , x = −5$]
]

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

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

#solutionbox[
no real solution
]

#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: "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: "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")[$9 y^(2) + 12 y + 4 = 9$]

#strong[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")[$r = ± #h(0.2em) 4$]
]

#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")[$a = ± #h(0.2em) 3 sqrt(2)$]
]

#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[
no real solution
]

#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")[$c = ± #h(0.2em) frac(5 sqrt(6), 6)$]
]

#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[
no real solution
]

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


==== Everyday Math

Paola has enough mulch to cover 48 square feet. She wants to use it to make three square vegetable gardens of equal sizes. Solve the equation #math.equation(block: false, alt: "3 s squared equals 48")[$3 s^(2) = 48$] to find #math.equation(block: false, alt: "s")[$s$], the length of each garden side.

#solutionbox[
4 feet
]

Kathy is drawing up the blueprints for a house she is designing. She wants to have four square windows of equal size in the living room, with a total area of 64 square feet. Solve the equation #math.equation(block: false, alt: "4 s squared equals 64")[$4 s^(2) = 64$] to find #math.equation(block: false, alt: "s")[$s$], the length of the sides of the windows.


==== Writing Exercises

Explain why the equation #math.equation(block: false, alt: "x squared plus 12 equals 8")[$x^(2) + 12 = 8$] has no solution.

#solutionbox[
Answers will vary.
]

Explain why the equation #math.equation(block: false, alt: "y squared plus 8 equals 12")[$y^(2) + 8 = 12$] has two solutions.


==== Self Check

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

#figure(figph[This table has three rows and four columns. The first row is a header row and it labels each column. The first column is labeled “I can …”, the second “Confidently”, the third “With some help” and the last “No–I don’t get it”. In the “I can…” column the next row reads “solve quadratic equations of the form a x squared equals k using the square root property.” and the last row reads “solve quadratic equations of the form a times the quantity x minus h squared equals k using the square root property.” The remaining columns are blank.], alt: "This table has three rows and four columns. The first row is a header row and it labels each column. The first column is labeled “I can …”, the second “Confidently”, the third “With some help” and the last “No–I don’t get it”. In the “I can…” column the next row reads “solve quadratic equations of the form a x squared equals k using the square root property.” and the last row reads “solve quadratic equations of the form a times the quantity x minus h squared equals k using the square root property.” The remaining columns are blank.", caption: none)

ⓑ 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.