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10.1: 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: 75.
  2. Simplify: 643
  3. Factor: 4x212x+9.

Quadratic equations are equations of the form ax2+bx+c=0, where a0. They differ from linear equations by including a term with the variable raised to the second power. 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 use three other methods to solve quadratic equations.

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=9Put the equation in standard form.x29=0Factor the left side.(x3)(x+3)=0Use the Zero Product Property.(x3)=0,(x+3)=0Solve each equation.x=3,x=3Combine the two solutions into±formx=±3

(The solution is read ‘x is equal to positive or negative 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. 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.

These equations are all of the form x2=k.
We defined the square root of a number in this way:

If n2=m, then n is a square root of m.

This leads to the Square Root Property.

Definition: SQUARE ROOT PROPERTY

If x2=k, and k0, then 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

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

x2=9Use the Square Root Property.x=±9Simplify the radical.x=±3Rewrite 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 x2=7.

Use the Square Root Property. x=±7Rewrite to show two solutions.x=7,x=7We cannot simplify7 so we leave the answer as a radical.

Example 10.1.1

Solve: x2=169

Answer

x2=169Use the Square Root Property.x=±169Simplify the radical.x=±13Rewrite to show two solutions.x=13,x=13

Example 10.1.2

Solve: x2=81

Answer

x=9, x=−9

Example 10.1.3

Solve: y2=121

Answer

y = 11, y = −11

How to Solve a Quadratic Equation of the Form ax2=k Using the Square Root Property

Example 10.1.4

Solve: x248=0

Answer
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.Step two is to use the Square Root Property to get x equals plus or minus the square root of 48.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.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.
Example 10.1.5

Solve: x250=0

Answer

x=52,x=52

Example 10.1.6

Solve: y227=0

Answer

y=33,x=33

Definition: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.

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.

Example 10.1.7

Solve: 5m2=80

Answer
The quadratic term is isolated. 5m2=80
Divide by 5 to make its cofficient 1. 5m25=805
Simplify. m2=16
Use the Square Root Property. m=±16
Simplify the radical. m=±4
Rewrite to show two solutions. m=4,m=−4

Check the solutions.
.

 
Example 10.1.8

Solve: 2x2=98.

Answer

x=7, x=−7

Example 10.1.9

Solve: 3z2=108.

Answer

z=6, z=−6

The Square Root Property started by stating, ‘If x2=k, and k0’. What will happen if k<0? This will be the case in the next example.

Example 10.1.10

Solve: q2+24=0.

Answer

q2=24Isolate the quadratic term.q2=24Use the Square Root Property.q=±24The24is not a real numberThere is no real solution

Example 10.1.11

Solve: c2+12=0.

Answer

no real solution

Example 10.1.12

Solve: d2+81=0.

Answer

no real solution

Remember, we first isolate the quadratic term and then make the coefficient equal to one.
Example 10.1.13

Solve: 23u2+5=17.

Answer
  23u2+5=17
Isolate the quadratic term.

23u2=12

Multiply by 32 to make the coefficient 1. 32·23u2=32·12
Simplify. u2=18
Use the Square Root Property. u=±18
Simplify the radical. u=±92
Simplify. u=±32
Rewrite to show two solutions. u=32, u=32

Check.
.

 
Example 10.1.14

Solve: 12x2+4=24

Answer

x=210, x=210

Example 10.1.15

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 10.1.16

Solve: 2c24=45.

Answer
  2c24=45
Isolate the quadratic term. 2c2=49
Divide by 2 to make the coefficient 1. 2c22=492
Simplify. c2=492
Use the Square Root Property. c=±492
Simplify the radical. c=±492
Rationalize the denominator. c=±49222
Simplify. c=±722
Rewrite to show two solutions. c=722, c=722
Check. We leave the check for you.  
Example 10.1.17

Solve: 5r22=34.

Answer

r=655, r=655

Example 10.1.18

Solve: 3t2+6=70.

Answer

t=833,t=833

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

We can use the Square Root Property to solve an equation like (x3)2=16, too. We will treat the whole binomial, (x−3), as the quadratic term.

Example 10.1.19

Solve: (x3)2=16.

Answer
  (x3)2=16
Use the Square Root Property. x3=±16
Simplify. x3=±4
Write as two equations. x3=4, x3=4
Solve. x=7,x=−1

Check.
.

 
Example 10.1.20

Solve: (q+5)2=1.

Answer

q=−6, q=−4

Example 10.1.21

Solve: (r3)2=25.

Answer

r=8, r=−2

Example 10.1.22

Solve: (y7)2=12.

Answer
  (y7)2=12.
Use the Square Root Property. y7=±12
Simplify the radical. y7=±23
Solve for y. y=7±23
Rewrite to show two solutions. y=7+23,y=723

Check.
.

 
Example 10.1.23

Solve: (a3)2=18.

Answer

a=3+32, a=332

Example 10.1.24

Solve: (b+2)2=40.

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 10.1.25

Solve: (x12)2=54.

Answer
  (x12)2=54
Use the Square Root Property. (x12)=±54
Rewrite the radical as a fraction of square roots. (x12)=±54
Simplify the radical. (x12)=±52
Solve for x. x=12+±52
Rewrite to show two solutions. x=12+52, x=1252
Check. We leave the check for you  
​​​​​​
Example 10.1.26

Solve: (x13)2=59.

Answer

x=13+53, x=1353

Example 10.1.27

Solve: (y34)2=716.

Answer

y=34+74, y=3474,

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

Example 10.1.28

Solve: (x2)2+3=30.

Answer
  (x2)2+3=30
Isolate the binomial term. (x2)2=27
Use the Square Root Property. x2=±27
Simplify the radical. x2=±33
Solve for x. x=2+±33
x2=±33 x=2+33, x=233
Check. We leave the check for you  
Example 10.1.29

Solve: (a5)2+4=24.

Answer

a=5+25, a=525

Example 10.1.30

Solve: (b3)28=24.

Answer

b=3+42, b=342

Example 10.1.31

Solve:(3v7)2=12.

Answer
  (3v7)2=12
Use the Square Root Property. 3v7=±12
The 12 is not a real number. There is no real solution.
Example 10.1.32

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

Answer

no real solution

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 10.1.33

Solve: p210p+25=18.

Answer

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

  p210p+25=18
Factor the perfect square trinomial. (p5)2=18
Use the Square Root Property. p5=±18
Simplify the radical. p5=±32
Solve for p. p=5±32
Rewrite to show two solutions. p=5+32, p=532
Check. We leave the check for you.  
Example 10.1.34

Solve: x26x+9=12.

Answer

x=3+23, x=323

Example 10.1.35

Solve: y2+12y+36=32.

Answer

y=6+42, y=642

Example 10.1.36

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

Answer

Again, 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±42

n=1±42

Rewrite to show two solutions. n=1+42, n=142
Simplify each equation. n=32, n=52

Check.
.

 
Example 10.1.37

Solve: 9m212m+4=25.

Answer

m=73, m=1

Example 10.1.38

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

Answer

n=34, n=74

Access these online resources for additional instruction and practice with solving quadratic equations:

Key Concepts

  • Square Root Property
    If x2=k, and k0, then x=k or x=k.

Glossary

quadratic equation
A quadratic equation is an equation of the form ax2+bx+c=0 where a0.
Square Root Property
The Square Root Property states that, if x2=k, and k0, then x=k or x=k.

This page titled 10.1: Solve Quadratic Equations Using the Square Root Property is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by OpenStax.

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