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11.2: Solve Quadratic Equations Using the Square Root Property

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Learning Objectives

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

  • Solve quadratic equations of the form ax2=k using the Square Root Property
  • Solve quadratic equations of the form a(xh)2=k using the Square Root Property

Before you get started, take this readiness quiz.

  1. Simplify: 128.
    If you missed this problem, review Example 8.13.
  2. Simplify: 325.
    If you missed this problem, review Example 8.50.
  3. Factor: 9x212x+4.
    If you missed this problem, review Example 6.23.

A quadratic equation is an equation of the form ax2+bx+c=0, where a0. 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 ax2=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.

x2=9

Put the equation in standard form.

x29=0

Factor the difference of squares.

(x3)(x+3)=0

Use the Zero Produce Property.

x3=0x3=0

Solve each equation.

x=3x=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 n2=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.

Definition 11.2.1

Square Root Property

If x2=k, then

x=k or x=k or x=±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=±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.

x2=9 Use the Square Root Property. x=±9x=±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.

x2=7

Use the Square Root Property. x=7,x=7

We cannot simplify 7, so we leave the answer as a radical.

Example 11.2.1 How to Solve a Quadratic Equation of the form ax2k Using the Square Root Property

Solve: x250=0.

Solution:

Step 1: Isolate the quadratic term and make its coefficient one. Add 50 to both sides to get x2 by itself. x250=0x2=50
Step 2: Use the Square Root Property. Remember to write the ± symbol. x=±50
Step 3: Simplify the radical. Rewrite to show two solutions. x=±252x=±52x=52,x=52
Step 4: Check the solutions. Substitute in x=52 and x=52

x250=0(52)250?=025250?=00=0

x250=0(52)250?=025250?=00=0

Table 9.1.1
Exercise 11.2.1

Solve: x248=0.

Answer

x=43,x=43

Exercise 11.2.2

Solve: y227=0.

Answer

y=33,y=33

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

Solve a Quadratic Equation Using the Square Root Property

  1. Isolate the quadratic term and make its coefficient one.
  2. Use Square Root Property.
  3. Simplify the radical.
  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 11.2.2

Solve: 3z2=108.

Solution:

  3z2=108
The quadratic term is isolated. Divide by 3 to make its coefficient 1. 3z23=1083
Simplify. z2=36
Use the Square Root Property. z=±36
Simplify the radical. z=±6
Rewrite to show two solutions. z=6,z=6

Check the solutions:

.
Figure 9.1.1
 
Table 9.1.2
Exercise 11.2.3

Solve: 2x2=98.

Answer

x=7,x=7

Exercise 11.2.4

Solve: 5m2=80.

Answer

m=4,m=4

The Square Root Property states ‘If x2=k,’ What will happen if k<0? This will be the case in the next example.

Example 11.2.3

Solve: x2+72=0.

Solution:

  x2+72=0
Isolate the quadratic term. x2=72
Use the Square Root Property. x=±72
Simplify using complex numbers. x=±72i
Simplify the radical. x=±62i
Rewrite to show two solutions x=62i,x=62i

Check the solutions:

x2+72=0(62i)2+72?=0(6)2(2)2i2+72?=0362(1)+72?=00=0

 

 

x2+72=0(62i)2+72?=0(6)2(2)2i2+72?=0362(1)+72?=00=0

Table 9.1.3
Exercise 11.2.5

Solve: c2+12=0.

Answer

c=23i,c=23i

Exercise 11.2.6

Solve: q2+24=0.

Answer

c=26i,c=26i

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 11.2.4

Solve: 23u2+5=17.

Solution:

  23u2+5=17
Isolate the quadratic term. .
Multiply by 32 to make the coefficient 1. .
Simplify. .
Use the Square Root Property. .
Simplify the radical. .
Simplify. .
Rewrite to show two solutions. .

Check:

23u2+5=1723(32)2+5?=172318+5?=1712+5?=1717=17

 

 

23u2+5=1723(32)2+5?=172318+5?=1712+5?=1717=17

Table 9.1.4
Exercise 11.2.7

Solve: 12x2+4=24.

Answer

x=210,x=210

Exercise 11.2.8

Solve: 34y23=18.

Answer

y=27,y=27

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

Example 11.2.5

Solve: 2x28=41.

Solution:

  .
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 te check for you.

 
Table 9.1.5
Exercise 11.2.9

Solve: 5r22=34.

Answer

r=655,r=655

Exercise 11.2.10

Solve: 3t2+6=70.

Answer

t=833,t=833

Solve Quadratic Equation of the Form a(xh)2=k Using the Square Root Property

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

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.
Figure 9.1.20

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 (xh)2.

Example 11.2.6

Solve: 4(y7)2=48.

Solution:

  4(y7)2=48
Divide both sides by the coefficient 4. (y7)2=12
Use the Square Root Property on the binomial. y7=±12
Simplify the radical. y7=±23
Solve for y. y=7±23
Rewrite to show two solutions. y=7+23
y=723

Check:

.
Figure 9.1.21
 
Table 9.1.6
Exercise 11.2.11

Solve: 3(a3)2=54.

Answer

a=3+32,a=332

Exercise 11.2.12

Solve: 2(b+2)2=80.

Answer

b=2+210,b=2210

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

Example 11.2.7

Solve: (x13)2=59.

Solution:

(x13)2=59

Use the Square Root Property.

x13=±59

Rewrite the radical as a fraction of square roots.

x13=±59

Simplify the radical.

x13=±53

Solve for x.

x=13±53

Rewrite to show two solutions.

x=13+53,x=1353

Check:

We leave the check for you.

Exercise 11.2.13

Solve: (x12)2=54.

Answer

x=12+52,x=1252

Exercise 11.2.14

Solve: (y+34)2=716.

Answer

y=34+74,y=3474

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

Example 11.2.8

Solve: 2(x2)2+3=57.

Solution:

2(x2)2+3=57

Subtract 3 from both sides to isolate the binomial term.

2(x2)2=54

Divide both sides by 2.

(x2)2=27

Use the Square Root Property.

x2=±27

Simplify the radical.

x2=±33

Solve for x.

x=2±33

Rewrite to show two solutions.

x=2+33,x=233

Check:

We leave the check for you.

Exercise 11.2.15

Solve: 5(a5)2+4=104.

Answer

a=5+25,a=525

Exercise 11.2.16

Solve: 3(b+3)28=88.

Answer

b=3+42,b=342

Sometimes the solutions are complex numbers.

Example 11.2.9

Solve: (2x3)2=12.

Solution:

(2x3)2=12

Use the Square Root Property.

2x3=±12

Simplify the radical.

2x3=±23i

Add 3 to both sides.

2x=3±23i

Divide both sides by 2.

x=3±23i2

Rewrite in standard form.

x=32±23i2

Simplify.

x=32±3i

Rewrite to show two solutions.

x=32+3i,x=323i

Check:

We leave the check for you.

Exercise 11.2.17

Solve: (3r+4)2=8.

Answer

r=43+22i3,r=4322i3

Exercise 11.2.18

Solve: (2t8)2=10.

Answer

t=4+10i2,t=410i2

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

Example 11.2.10

Solve: 4n2+4n+1=16.

Solution:

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

  4n2+4n+1=16
Factor the perfect square trinomial. (2n+1)2=16
Use the Square Root Property. 2n+1=±16
Simplify the radical. 2n+1=±4
Solve for n. 2n=1±4
Divide each side by 2. 2n2=1±42n=1±42
Rewrite to show two solutions. n=1+42,n=142
Simplify each equation. n=32,n=52

Check:

.
Figure 9.1.22
 
Table 9.1.7
Exercise 11.2.19

Solve: 9m212m+4=25.

Answer

m=73,m=1

Exercise 11.2.20

Solve: 16n2+40n+25=4.

Answer

n=34,n=74

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

Key Concepts

  • Square Root Property
    • If x2=k, then x=k or x=kor x=±k
    How to solve a quadratic equation using the square root property.
    1. Isolate the quadratic term and make its coefficient one.
    2. Use Square Root Property.
    3. Simplify the radical.
    4. Check the solutions.

This page titled 11.2: Solve Quadratic Equations Using the Square Root Property is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Chau D Tran.

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