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# 10.4: Use Square Roots to Solve Quadratic Equations

Difficulty Level: At Grade Created by: CK-12
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What if you had a quadratic equation like $4x^2 - 9 = 0$ in which both terms were perfect squares? How could you solve such an equation? After completing this Concept, you'll be able to solve quadratic equations like this one that involve perfect squares.

### Guidance

So far you know how to solve quadratic equations by factoring. However, this method works only if a quadratic polynomial can be factored. In the real world, most quadratics can’t be factored, so now we’ll start to learn other methods we can use to solve them. In this Concept, we’ll examine equations in which we can take the square root of both sides of the equation in order to arrive at the result.

Solve Quadratic Equations Involving Perfect Squares

Let’s first examine quadratic equations of the type

$x^2 - c = 0$

We can solve this equation by isolating the $x^2$ term: $x^2 = c$

Once the $x^2$ term is isolated we can take the square root of both sides of the equation. Remember that when we take the square root we get two answers: the positive square root and the negative square root:

$x = \sqrt{c} \qquad \text{and} \qquad x = -\sqrt{c}$

Often this is written as $x = \pm \sqrt{c}$ .

#### Example A

a) $x^2 - 4 = 0$

b) $x^2 - 25 = 0$

Solution

a) $x^2 - 4 = 0$

Isolate the $x^2$ : $x^2 = 4$

Take the square root of both sides: $x = \sqrt{4}$ and $x = - \sqrt{4}$

The solutions are $x = 2$ and $x = -2$ .

b) $x^2 - 25 = 0$

Isolate the $x^2$ : $x^2 = 25$

Take the square root of both sides: $x = \sqrt{25}$ and $x = - \sqrt{25}$

The solutions are $x = 5$ and $x = -5$ .

We can also find the solution using the square root when the $x^2$ term is multiplied by a constant—in other words, when the equation takes the form

$ax^2 - c = 0$

We just have to isolate the $x^2$ :

$ax^2 & = b\\x^2 & = \frac{b}{a}$

Then we can take the square root of both sides of the equation:

$x = \sqrt{\frac{b}{a}} \qquad \text{and} \qquad x = - \sqrt{ \frac{b}{a}}$

Often this is written as: $x = \pm \sqrt{\frac{b}{a}}$ .

#### Example B

a) $9x^2 - 16 = 0$

b) $81x^2 - 1 = 0$

Solution

a) $9x^2 - 16 = 0$

Isolate the $x^2$ :

$9x^2 & = 16\\x^2 & = \frac{16}{9}$

Take the square root of both sides: $x = \sqrt{\frac{16}{9}}$ and $x = - \sqrt{ \frac{16}{9}}$

Answer: $x = \frac{4}{3}$ and $x = - \frac{4}{3}$

b) $81x^2 - 1 = 0$

Isolate the $x^2$ :

$81x^2 & = 1\\x^2 & = \frac{1}{81}$ Take the square root of both sides: $x = \sqrt{\frac{1}{81}}$ and $x = - \sqrt{ \frac{1}{81}}$

Answer: $x = \frac{1}{9}$ and $x = - \frac{1}{9}$

As you’ve seen previously, some quadratic equations have no real solutions.

#### Example C

a) $x^2 + 1 = 0$

b) $4x^2 + 9 = 0$

Solution

a) $x^2 + 1 = 0$

Isolate the $x^2$ : $x^2 = -1$

Take the square root of both sides: $x = \sqrt{-1}$ and $x = - \sqrt{-1}$

Square roots of negative numbers do not give real number results, so there are no real solutions to this equation.

b) $4x^2 + 9 = 0$

Isolate the $x^2$ :

$4x^2 & = -9\\x^2 & = - \frac{9}{4}$ Take the square root of both sides: $x = \sqrt{ - \frac{9}{4}}$ and $x = - \sqrt{ - \frac{9}{4}}$

There are no real solutions.

We can also use the square root function in some quadratic equations where both sides of an equation are perfect squares. This is true if an equation is of this form:

$(x - 2)^2 = 9$

Both sides of the equation are perfect squares. We take the square root of both sides and end up with two equations: $x - 2 = 3$ and $x - 2 = -3$ .

Solving both equations gives us $x = 5$ and $x = -1$ .

#### Example D

a) $(x - 1)^2 = 4$

b) $(x + 3)^2 = 1$

Solution

a) $(x - 1)^2 = 4$

$\text{Take the square root of both sides}: & & x - 1 & = 2 \ \text{and} \ x - 1 = -2\\\text{Solve each equation}: & & x & = 3 \ \text{and} \ x = -1$

Answer: $x =3$ and $x = -1$

b) $(x + 3)^2 = 1$

$\text{Take the square root of both sides}: & & x + 3 & = 1 \ \text{and} \ x + 3 = -1\\\text{Solve each equation}: & & x & = -2 \ \text{and} \ x = -4$

Answer: $x = -2$ and $x = -4$

It might be necessary to factor the right-hand side of the equation as a perfect square before applying the method outlined above.

Watch this video for help with the Examples above.

### Vocabulary

• The solutions of a quadratic equation are often called the roots or zeros.

### Guided Practice

a) $x^2 + 8x + 16 = 25$

b) $4x^2 - 40x + 25 = 9$

Solution

a) $x^2 + 8x + 16 = 25$

$& \text{Factor the right-hand-side}: & & x^2 + 8x + 16 = (x + 4)^2 \quad \text{so} \quad (x + 4)^2 = 25\\& \text{Take the square root of both sides}: & & x + 4 = 5 \ \text{and} \ x + 4 = -5 \\& \text{Solve each equation}: & & x = 1 \ \text{and} \ x = -9$

Answer: $x = 1$ and $x = -9$

b) $4x^2 - 20x + 25 = 9$

$& \text{Factor the right-hand-side}: & & 4x^2 - 20x + 25 = (2x - 5)^2 \quad \text{so} \quad (2x - 5)^2 = 9\\& \text{Take the square root of both sides}: & & 2x - 5 = 3 \ \text{and} \ 2x - 5 = -3 \\& \text{Solve each equation}: & & 2x = 8 \ \text{and} \ 2x = 2$

Answer: $x = 4$ and $x =1$

### Practice

1. $x^2 - 1 = 0$
2. $x^2 - 100 = 0$
3. $x^2 + 16 = 0$
4. $9x^2 - 1 = 0$
5. $4x^2 - 49 = 0$
6. $64x^2 - 9 = 0$
7. $x^2 - 81 = 0$
8. $25x^2 - 36 = 0$
9. $x^2 + 9 = 0$
10. $x^2 - 16 = 0$
11. $x^2 - 36 = 0$
12. $16x^2 - 49 = 0$
13. $(x - 2)^2 = 1$
14. $(x + 5)^2 = 16$
15. $(2x - 1)^2 - 4 = 0$
16. $(3x + 4)^2 = 9$
17. $(x - 3)^2 + 25 = 0$
18. $x^2 - 10x + 25 =9$
19. $x^2 + 18x + 81 = 1$
20. $4x^2 - 12x + 9 = 16$
21. $2(x + 3)^2 = 8$

Oct 01, 2012

Oct 28, 2014