9.2: Solve Quadratic Equations Using the Square Root Property (2024)

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Learning Objectives

By the end of this section, you will be able to:

Solve quadratic equations of the form ${a x}^{2} = k$ using the Square Root Property
Solve quadratic equations of the form $a {(x - h)}^{2} = k$ using the Square Root Property

Be Prepared 9.1

Before you get started, take this readiness quiz.

Simplify: $\sqrt{128} .$
If you missed this problem, review Example 8.13.

Be Prepared 9.2

Simplify: $\sqrt{\frac{32}{5}}$ .
If you missed this problem, review Example 8.50.

Be Prepared 9.3

Factor: $9 x^{2} - 12 x + 4$ .
If you missed this problem, review Example 6.23.

A quadratic equation is an equation of the form ax² + bx + c = 0, where $a \neq 0$ . Quadratic equations differ from linear equations by including a quadratic term with the variable raised to the second power of the form ax². 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 $a x^{2} = k$ 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 x² = 9.

	$x^{2} = 9$
Put the equation in standard form.	$x^{2} - 9 = 0$
Factor the difference of squares.	$(x - 3) (x + 3) = 0$
Use the Zero Product Property.	$x - 3 = 0 x - 3 = 0$
Solve each equation.	$x = 3 x = −3$

We can easily use factoring to find the solutions of similar equations, like x² = 16 and x² = 25, because 16 and 25 are perfect squares. In each case, we would get two solutions, $x = 4, x = −4$ and $x = 5, x = −5 .$

But what happens when we have an equation like x² = 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 square root of 169. Also, (−13)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:

$If n^{2} = m, then n is a square root of m .$

Since these equations are all of the form x² = k, the square root definition tells us the solutions are the two square roots of k. This leads to the Square Root Property.

Square Root Property

If x² = k, then

$x = \sqrt{k} or x = − \sqrt{k} or x = \pm \sqrt{k} .$

Notice that the Square Root Property gives two solutions to an equation of the form x² = k, the principal square root of $k$ and its opposite. We could also write the solution as $x = \pm \sqrt{k} .$ We read this as x equals positive or negative the square root of k.

Now we will solve the equation x² = 9 again, this time using the Square Root Property.

	$x^{2} = 9$
Use the Square Root Property.	$x = \pm \sqrt{9}$
	$x = \pm 3$
	$So x = 3 or x = −3 .$

What happens when the constant is not a perfect square? Let’s use the Square Root Property to solve the equation x² = 7.

	$x^{2} = 7$
Use the Square Root Property.	$x = \sqrt{7}, x = − \sqrt{7}$

We cannot simplify $\sqrt{7}$ , so we leave the answer as a radical.

Example 9.1

How to solve a Quadratic Equation of the form ax² = k Using the Square Root Property

Solve: $x^{2} - 50 = 0 .$

Answer

Try It 9.1

Solve: $x^{2} - 48 = 0 .$

Try It 9.2

Solve: $y^{2} - 27 = 0 .$

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

How To

Solve a quadratic equation using the square root property.

Step 1. Isolate the quadratic term and make its coefficient one.
Step 2. Use Square Root Property.
Step 3. Simplify the radical.
Step 4. 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.

Example 9.2

Solve: $3 z^{2} = 108 .$

Answer

	$3 z^{2} = 108$
The quadratic term is isolated. Divide by 3 to make its coefficient 1.	$\frac{3 z^{2}}{3} = \frac{108}{3}$
Simplify.	$z^{2} = 36$
Use the Square Root Property.	$z = \pm \sqrt{36}$
Simplify the radical.	$z = \pm 6$
Rewrite to show two solutions.	$z = 6, z = −6$
Check the solutions:

Try It 9.3

Solve: $2 x^{2} = 98 .$

Try It 9.4

Solve: $5 m^{2} = 80 .$

The Square Root Property states ‘If $x^{2} = k$ ,’ What will happen if $k < 0 ?$ This will be the case in the next example.

Example 9.3

Solve: $x^{2} + 72 = 0$ .

Answer

	$x^{2} + 72 = 0$
Isolate the quadratic term.	$x^{2} = −72$
Use the Square Root Property.	$x = \pm \sqrt{−72}$
Simplify using complex numbers.	$x = \pm \sqrt{72} i$
Simplify the radical.	$x = \pm 6 \sqrt{2} i$
Rewrite to show two solutions.	$x = 6 \sqrt{2} i, x = −6 \sqrt{2} i$
Check the solutions:

Try It 9.5

Solve: $c^{2} + 12 = 0 .$

Try It 9.6

Solve: $q^{2} + 24 = 0 .$

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.

Example 9.4

Solve: $\frac{2}{3} u^{2} + 5 = 17 .$

Answer

	$\frac{2}{3} u^{2} + 5 = 17$
Isolate the quadratic term.
Multiply by $\frac{3}{2}$ to make the coefficient 1.
Simplify.
Use the Square Root Property.
Simplify the radical.
Simplify.
Rewrite to show two solutions.
Check:

Try It 9.7

Solve: $\frac{1}{2} x^{2} + 4 = 24 .$

Try It 9.8

Solve: $\frac{3}{4} y^{2} - 3 = 18 .$

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

Example 9.5

Solve: $2 x^{2} - 8 = 41 .$

Answer


Isolate the quadratic term.
Divide by $2$ to make the coefficient 1.
Simplify.
Use the Square Root Property.
Rewrite the radical as a fraction of square roots.
Rationalize the denominator.
Simplify.
Rewrite to show two solutions.
Check: We leave the check for you.

Try It 9.9

Solve: $5 r^{2} - 2 = 34 .$

Try It 9.10

Solve: $3 t^{2} + 6 = 70 .$

Solve Quadratic Equations of the Form a(x − h)² = k Using the Square Root Property

We can use the Square Root Property to solve an equation of the form a(x − h)² = k as well. Notice that the quadratic term, x, in the original form ax² = k is replaced with (x − h).

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 a, then the Square Root Property can be used on (x − h)².

Example 9.6

Solve: $4 {(y - 7)}^{2} = 48 .$

Answer

	$4 {(y - 7)}^{2} = 48$
Divide both sides by the coefficient 4.	${(y - 7)}^{2} = 12$
Use the Square Root Property on the binomial	$y - 7 = \pm \sqrt{12}$
Simplify the radical.	$y - 7 = \pm 2 \sqrt{3}$
Solve for $y .$	$y = 7 \pm 2 \sqrt{3}$
Rewrite to show two solutions.	$y = 7 + 2 \sqrt{3},$ $y = 7 - 2 \sqrt{3}$
Check:

Try It 9.11

Solve: $3 {(a - 3)}^{2} = 54 .$

Try It 9.12

Solve: $2 {(b + 2)}^{2} = 80 .$

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

Example 9.7

Solve: ${(x - \frac{1}{3})}^{2} = \frac{5}{9} .$

Answer

	${(x - \frac{1}{3})}^{2} = \frac{5}{9}$
Use the Square Root Property.	$x - \frac{1}{3} = \pm \sqrt{\frac{5}{9}}$
Rewrite the radical as a fraction of square roots.	$x - \frac{1}{3} = \pm \frac{\sqrt{5}}{\sqrt{9}}$
Simplify the radical.	$x - \frac{1}{3} = \pm \frac{\sqrt{5}}{3}$
Solve for $x$ .	$x = \frac{1}{3} \pm \frac{\sqrt{5}}{3}$
Rewrite to show two solutions.	$x = \frac{1}{3} + \frac{\sqrt{5}}{3}, x = \frac{1}{3} - \frac{\sqrt{5}}{3}$
Check: We leave the check for you.

Try It 9.13

Solve: ${(x - \frac{1}{2})}^{2} = \frac{5}{4} .$

Try It 9.14

Solve: ${(y + \frac{3}{4})}^{2} = \frac{7}{16} .$

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

Example 9.8

Solve: $2 {(x - 2)}^{2} + 3 = 57 .$

Answer

	$2 {(x - 2)}^{2} + 3 = 57$
Subtract 3 from both sides to isolate the binomial term.	$2 {(x - 2)}^{2} = 54$
Divide both sides by 2.	${(x - 2)}^{2} = 27$
Use the Square Root Property.	$x - 2 = \pm \sqrt{27}$
Simplify the radical.	$x - 2 = \pm 3 \sqrt{3}$
Solve for $x$ .	$x = 2 \pm 3 \sqrt{3}$
Rewrite to show two solutions.	$x = 2 + 3 \sqrt{3}, x = 2 - 3 \sqrt{3}$
Check: We leave the check for you.

Try It 9.15

Solve: $5 {(a - 5)}^{2} + 4 = 104 .$

Try It 9.16

Solve: $3 {(b + 3)}^{2} - 8 = 88 .$

Sometimes the solutions are complex numbers.

Example 9.9

Solve: ${(2 x - 3)}^{2} = −12 .$

Answer

	${(2 x - 3)}^{2} = −12$
Use the Square Root Property.	$2 x - 3 = \pm \sqrt{−12}$
Simplify the radical.	$2 x - 3 = \pm 2 \sqrt{3} i$
Add 3 to both sides.	$2 x = 3 \pm 2 \sqrt{3} i$
Divide both sides by 2.	$x = \frac{3 \pm 2 \sqrt{3} i}{2}$
Rewrite in standard form.	$x = \frac{3}{2} \pm \frac{2 \sqrt{3} i}{2}$
Simplify.	$x = \frac{3}{2} \pm \sqrt{3} i$
Rewrite to show two solutions.	$x = \frac{3}{2} + \sqrt{3} i, x = \frac{3}{2} - \sqrt{3} i$
Check: We leave the check for you.

Try It 9.17

Solve: ${(3 r + 4)}^{2} = −8 .$

Try It 9.18

Solve: ${(2 t - 8)}^{2} = −10 .$

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

Example 9.10

Solve: $4 n^{2} + 4 n + 1 = 16 .$

Answer

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

	$4 n^{2} + 4 n + 1 = 16$
Factor the perfect square trinomial.	${(2 n + 1)}^{2} = 16$
Use the Square Root Property.	$2 n + 1 = \pm \sqrt{16}$
Simplify the radical.	$2 n + 1 = \pm 4$
Solve for $n$ .	$2 n = −1 \pm 4$
Divide each side by 2.	$\frac{2 n}{2} = \frac{−1 \pm 4}{2}$ $n = \frac{−1 \pm 4}{2}$
Rewrite to show two solutions.	$n = \frac{−1 + 4}{2}$ , $n = \frac{−1 - 4}{2}$
Simplify each equation.	$n = \frac{3}{2}$ , $n = - \frac{5}{2}$
Check:

Try It 9.19

Solve: $9 m^{2} - 12 m + 4 = 25 .$

Try It 9.20

Solve: $16 n^{2} + 40 n + 25 = 4 .$

Media

Access this online resource for additional instruction and practice with using the Square Root Property to solve quadratic equations.

Section 9.1 Exercises

Practice Makes Perfect

Solve Quadratic Equations of the Form ax² = k Using the Square Root Property

In the following exercises, solve each equation.

$a^{2} = 49$

$b^{2} = 144$

$r^{2} - 24 = 0$

$t^{2} - 75 = 0$

$u^{2} - 300 = 0$

$v^{2} - 80 = 0$

$4 m^{2} = 36$

$3 n^{2} = 48$

$\frac{4}{3} x^{2} = 48$

10.

$\frac{5}{3} y^{2} = 60$

11.

$x^{2} + 25 = 0$

12.

$y^{2} + 64 = 0$

13.

$x^{2} + 63 = 0$

14.

$y^{2} + 45 = 0$

15.

$\frac{4}{3} x^{2} + 2 = 110$

16.

$\frac{2}{3} y^{2} - 8 = −2$

17.

$\frac{2}{5} a^{2} + 3 = 11$

18.

$\frac{3}{2} b^{2} - 7 = 41$

19.

$7 p^{2} + 10 = 26$

Mixed Practice

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

47.

$2 r^{2} = 32$

48.

$4 t^{2} = 16$

49.

${(a - 4)}^{2} = 28$

50.

${(b + 7)}^{2} = 8$

51.

$9 w^{2} - 24 w + 16 = 1$

52.

$4 z^{2} + 4 z + 1 = 49$

53.

$a^{2} - 18 = 0$

54.

$b^{2} - 108 = 0$

55.

${(p - \frac{1}{3})}^{2} = \frac{7}{9}$

56.

${(q - \frac{3}{5})}^{2} = \frac{3}{4}$

57.

$m^{2} + 12 = 0$

58.

$n^{2} + 48 = 0 .$

59.

$u^{2} - 14 u + 49 = 72$

60.

$v^{2} + 18 v + 81 = 50$

61.

${(m - 4)}^{2} + 3 = 15$

62.

${(n - 7)}^{2} - 8 = 64$

63.

${(x + 5)}^{2} = 4$

64.

${(y - 4)}^{2} = 64$

65.

$6 c^{2} + 4 = 29$

66.

$2 d^{2} - 4 = 77$

67.

${(x - 6)}^{2} + 7 = 3$

68.

${(y - 4)}^{2} + 10 = 9$

Writing Exercises

69.

In your own words, explain the Square Root Property.

70.

In your own words, explain how to use the Square Root Property to solve the quadratic equation ${(x + 2)}^{2} = 16$ .

Self Check

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

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:

…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.

…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?

…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.

9.2: Solve Quadratic Equations Using the Square Root Property (2024)

FAQs

How do you solve a quadratic equation using the square root property? ›

To solve an equation by using the square root property, you will first isolate the term that contains the squared variable. You can then take the square root of both sides and solve for the variable. Make sure to write the final answer in simplified form.

Tell Me More ›

What is the √ property? ›

It just so happens that there is a property we can use to solve these specific types of equations, and that property is called the square root property. The square root property can be used to solve certain quadratic equations, and it states that if x2 = c, then x = √c or x = -√c, where c is a number.

See Details ›

What is the square root formula for a quadratic equation? ›

Important Formulas on Roots of Quadratic Equations:

For a quadratic equation ax² + bx + c = 0, The roots are calculated using the formula, x = (-b ± √ (b² - 4ac) )/2a. Discriminant is, D = b² - 4ac.

Get More Info Here ›

What is the formula for the square root properties? ›

The square root is an inverse method of squaring a number. Hence, squares and square roots are related concepts. Suppose x is the square root of y, then it is represented as x=√y, or we can express the same equation as x² = y. Here, '√' is the radical symbol used to represent the root of numbers.

Keep Reading ›

How to solve equations with square roots? ›

Solving square-root equations

Step 1: Isolate the radical term.
Step 2: Take the square of both sides.
Step 3: Check for extraneous solutions.

View Details ›

What is a property example? ›

Key Takeaways. Property is any item that a person or a business has legal title over. Property can be tangible items, such as houses, cars, or appliances, or it can refer to intangible items that carry the promise of future worth, such as stock and bond certificates.

Keep Reading ›

What are 4 examples of property? ›

Properties

Commutativeexample. a + b = b + a2 + 6 = 6 + 2. ...
Associativeexample. (a + b) + c = a + ( b + c ) (1 + 6) + 3 = 1 + (6 + 3) ...
Distributiveexample. a × (b + c) = ab + ac3 × (6+2) = 3 × 6 + 3 × 2. ...
Closureexample. ...
Identityexample. ...
a + (−a ) = 06 + (−6) = 0. ...
Zero Productexample. ...
−1 × (−a) = −(−a) = a−1 × (−5) = −(−5) = 5.

Read On ›

How do you calculate property? ›

Methods of Property Valuation in India

Belt Method. This is the most common method of evaluating a large property's market value in urban areas. ...
Land and Building Method. ...
Comparative Property Valuation Method. ...
4. Development Method. ...
Guidance Value Method. ...
Fair Market Value. ...
External Factor. ...
Internal Factor.

More items...

May 10, 2024

Read The Full Story ›

How do you solve quadratic equations by square method? ›

Steps to Solving Equations by Completing the Square

Rewrite the equation in the form x² + bx = c.
Add to both sides the term needed to complete the square.
Factor the perfect square trinomial.
Solve the resulting equation by using the square root property.

How do you solve a quadratic equation by square root property? ›

In summary, when there is no linear term in a quadratic equation, one method to solve it is to use the square root property. In this approach, the x2 term (or more generally the squared term) is isolated first, and then the square root of both sides of the equal sign is taken.

Read The Full Story ›

What are the steps for the square root method? ›

Step 1: Divide the given number by divisor by identifying the suitable integer. Step 2: Multiply the divisor and integer (quotient) to get the number to be subtracted from the dividend. Step 3: Subtract the number from the dividend. Step 4: Bring down the remainder and another digit (if any) from the dividend.

Know More ›

How to find a quadratic equation from roots? ›

Imagine there are two different roots of a quadratic equation given as α and β then x2 – (α + β)x + αβ = 0 will be the formula to build the quadratic equation. Imagine the quadratic equation's standard form is ax2 + bx + c = 0 where a, b, and c are real numbers.

Get More Info ›

What is the root property of quadratic equation? ›

Answer: You must compute the discriminant to determine the number of roots in a quadratic ax²+bx+c=0 (b²-4ac). The quadratic has no real roots if the discriminant is less than 0. The quadratic has equal roots if the discriminant is equal to zero. There are two separate roots if the discriminant is greater than zero.

Can all quadratic equations be solved by the square root method? ›

Not all quadratic equations are solved by immediately taking the square root. Sometimes we have to isolate the squared term before taking its root. For example, to solve the equation 2 x 2 + 3 = 131 ‍ we should first isolate ‍ .

Show Me More ›

How do you find the roots of a quadratic using completing the square? ›

An alternative method to solve a quadratic equation is to complete the square. To solve an equation of the form x 2 + b x + c = 0 , consider the expression ( x + b 2 ) 2 + c . This can be rearranged to give ( x + b 2 ) 2 = ( b 2 ) 2 − c which can then be solved by taking the square root of both sides.

Get More Info Here ›

9.2: Solve Quadratic Equations Using the Square Root Property (2024)

Learning Objectives

Be Prepared 9.1

Be Prepared 9.2

Be Prepared 9.3

Solve Quadratic Equations of the form ax2=kax2=k using the Square Root Property

Square Root Property

Example 9.1

How to solve a Quadratic Equation of the form ax2 = k Using the Square Root Property

Try It 9.1

Try It 9.2

How To

Solve a quadratic equation using the square root property.

Example 9.2

Try It 9.3

Try It 9.4

Example 9.3

Try It 9.5

Try It 9.6

Example 9.4

Try It 9.7

Try It 9.8

Example 9.5

Try It 9.9

Try It 9.10

Solve Quadratic Equations of the Form a(x − h)2 = k Using the Square Root Property

Example 9.6

Try It 9.11

Try It 9.12

Example 9.7

Try It 9.13

Try It 9.14

Example 9.8

Try It 9.15

Try It 9.16

Example 9.9

Try It 9.17

Try It 9.18

Example 9.10

Try It 9.19

Try It 9.20

Media

Section 9.1 Exercises

Practice Makes Perfect

Mixed Practice

Writing Exercises

Self Check

FAQs

How do you solve a quadratic equation using the square root property? ›

How do you solve a quadratic equation by square root property? ›

References

Solve Quadratic Equations of the form $a x^{2} = k$ using the Square Root Property

How to solve a Quadratic Equation of the form ax² = k Using the Square Root Property

Solve Quadratic Equations of the Form a(x − h)² = k Using the Square Root Property