Precalculus Examples

x^{2} + 4 x + 4 = 0

$x^{2} + 4 x + 4 = 0$

Step 1

Use the quadratic formula to find the solutions.

\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 2

Substitute the values

a = 1

$a = 1$ ,

b = 4

$b = 4$ , and

c = 4

$c = 4$ into the quadratic formula and solve for

x

$x$ .

\frac{- 4 \pm \sqrt{4^{2} - 4 \cdot (1 \cdot 4)}}{2 \cdot 1}

$\frac{- 4 \pm \sqrt{4^{2} - 4 \cdot (1 \cdot 4)}}{2 \cdot 1}$

Step 3

Simplify.

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Step 3.1

Simplify the numerator.

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Step 3.1.1

Raise

4

$4$ to the power of

2

$2$ .

x = \frac{- 4 \pm \sqrt{16 - 4 \cdot 1 \cdot 4}}{2 \cdot 1}

$x = \frac{- 4 \pm \sqrt{16 - 4 \cdot 1 \cdot 4}}{2 \cdot 1}$

Step 3.1.2

Multiply

- 4 \cdot 1 \cdot 4

$- 4 \cdot 1 \cdot 4$ .

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Step 3.1.2.1

Multiply

- 4

$- 4$ by

1

$1$ .

x = \frac{- 4 \pm \sqrt{16 - 4 \cdot 4}}{2 \cdot 1}

$x = \frac{- 4 \pm \sqrt{16 - 4 \cdot 4}}{2 \cdot 1}$

Step 3.1.2.2

Multiply

- 4

$- 4$ by

4

$4$ .

x = \frac{- 4 \pm \sqrt{16 - 16}}{2 \cdot 1}

$x = \frac{- 4 \pm \sqrt{16 - 16}}{2 \cdot 1}$

x = \frac{- 4 \pm \sqrt{16 - 16}}{2 \cdot 1}

$x = \frac{- 4 \pm \sqrt{16 - 16}}{2 \cdot 1}$

Step 3.1.3

Subtract

16

$16$ from

16

$16$ .

x = \frac{- 4 \pm \sqrt{0}}{2 \cdot 1}

$x = \frac{- 4 \pm \sqrt{0}}{2 \cdot 1}$

Step 3.1.4

Rewrite

0

$0$ as

0^{2}

$0^{2}$ .

x = \frac{- 4 \pm \sqrt{0^{2}}}{2 \cdot 1}

$x = \frac{- 4 \pm \sqrt{0^{2}}}{2 \cdot 1}$

Step 3.1.5

Pull terms out from under the radical, assuming positive real numbers.

x = \frac{- 4 \pm 0}{2 \cdot 1}

$x = \frac{- 4 \pm 0}{2 \cdot 1}$

Step 3.1.6

- 4

$- 4$ plus or minus

0

$0$ is

- 4

$- 4$ .

x = \frac{- 4}{2 \cdot 1}

$x = \frac{- 4}{2 \cdot 1}$

x = \frac{- 4}{2 \cdot 1}

$x = \frac{- 4}{2 \cdot 1}$

Step 3.2

Multiply

2

$2$ by

1

$1$ .

x = \frac{- 4}{2}

$x = \frac{- 4}{2}$

Step 3.3

Divide

- 4

$- 4$ by

2

$2$ .

x = - 2

$x = - 2$

x = - 2

$x = - 2$

Step 4

The final answer is the combination of both solutions.

x = - 2

$x = - 2$ Double roots