# 10.4: Use Square Roots to Solve Quadratic Equations

**At Grade**Created by: CK-12

**Practice**Use Square Roots to Solve Quadratic Equations

What if you had a quadratic equation like

### Watch This

CK-12 Foundation: 1004S Solving Quadratic Equations Using Square Roots

### 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

We can solve this equation by isolating the

Once the

Often this is written as

#### Example A

*Solve the following quadratic equations:*

a)

b)

**Solution**

a)

Isolate the

Take the square root of both sides:

The solutions are **and**

b)

Isolate the

Take the square root of both sides:

The solutions are **and x=−5.**

We can also find the solution using the square root when the

We just have to isolate the

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

Often this is written as:

#### Example B

*Solve the following quadratic equations.*

a)

b)

**Solution**

a)

Isolate the

Take the square root of both sides:

**Answer:**

b)

Isolate the

**Answer:**

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

#### Example C

*Solve the following quadratic equations.*

a)

b)

**Solution**

a)

Isolate the

Take the square root of both sides: \begin{align*} x = \sqrt{-1}\end{align*} and \begin{align*}x = - \sqrt{-1}\end{align*}

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

b) \begin{align*} 4x^2 + 9 = 0\end{align*}

Isolate the \begin{align*}x^2\end{align*}:

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

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:

\begin{align*}(x - 2)^2 = 9\end{align*}

Both sides of the equation are perfect squares. We take the square root of both sides and end up with two equations: \begin{align*}x - 2 = 3\end{align*} and \begin{align*} x - 2 = -3\end{align*}.

Solving both equations gives us \begin{align*}x = 5\end{align*} and \begin{align*}x = -1\end{align*}.

#### Example D

*Solve the following quadratic equations.*

a) \begin{align*}(x - 1)^2 = 4\end{align*}

b) \begin{align*}(x + 3)^2 = 1\end{align*}

**Solution**

a) \begin{align*}(x - 1)^2 = 4\end{align*}

\begin{align*}\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\end{align*}

**Answer:** \begin{align*}x =3\end{align*} and \begin{align*}x = -1\end{align*}

b) \begin{align*}(x + 3)^2 = 1\end{align*}

\begin{align*}\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\end{align*}

**Answer:** \begin{align*}x = -2\end{align*} and \begin{align*}x = -4\end{align*}

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.

CK-12 Foundation: 1004 Solving Quadratic Equations Using Square Roots

### Vocabulary

- The
**solutions of a quadratic equation**are often called theor*roots**zeros.*

### Guided Practice

*Solve the following quadratic equations.*

a) \begin{align*}x^2 + 8x + 16 = 25\end{align*}

b) \begin{align*}4x^2 - 40x + 25 = 9\end{align*}

**Solution**

a) \begin{align*}x^2 + 8x + 16 = 25\end{align*}

\begin{align*}& \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 \end{align*}

**Answer:** \begin{align*}x = 1\end{align*} and \begin{align*}x = -9\end{align*}

b) \begin{align*}4x^2 - 20x + 25 = 9\end{align*}

\begin{align*}& \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 \end{align*}

**Answer:** \begin{align*}x = 4\end{align*} and \begin{align*} x =1\end{align*}

### Practice

Solve the following quadratic equations.

- \begin{align*}x^2 - 1 = 0\end{align*}
- \begin{align*}x^2 - 100 = 0\end{align*}
- \begin{align*}x^2 + 16 = 0\end{align*}
- \begin{align*}9x^2 - 1 = 0\end{align*}
- \begin{align*}4x^2 - 49 = 0\end{align*}
- \begin{align*}64x^2 - 9 = 0\end{align*}
- \begin{align*}x^2 - 81 = 0\end{align*}
- \begin{align*}25x^2 - 36 = 0\end{align*}
- \begin{align*}x^2 + 9 = 0\end{align*}
- \begin{align*}x^2 - 16 = 0\end{align*}
- \begin{align*}x^2 - 36 = 0\end{align*}
- \begin{align*}16x^2 - 49 = 0\end{align*}
- \begin{align*}(x - 2)^2 = 1\end{align*}
- \begin{align*}(x + 5)^2 = 16\end{align*}
- \begin{align*}(2x - 1)^2 - 4 = 0\end{align*}
- \begin{align*}(3x + 4)^2 = 9\end{align*}
- \begin{align*}(x - 3)^2 + 25 = 0\end{align*}
- \begin{align*}x^2 - 10x + 25 =9\end{align*}
- \begin{align*}x^2 + 18x + 81 = 1\end{align*}
- \begin{align*}4x^2 - 12x + 9 = 16\end{align*}
- \begin{align*}2(x + 3)^2 = 8\end{align*}

### Notes/Highlights Having trouble? Report an issue.

Color | Highlighted Text | Notes | |
---|---|---|---|

Show More |

### Image Attributions

Here you'll learn how to solve quadratic equations in which finding the solutions involves square roots.