Precalculus Examples

${(n + 2)}^{2} - 4 (2 n + 1) = 0$

Step 1

Simplify the left side.

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Step 1.1

Simplify each term.

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Step 1.1.1

Apply the distributive property.

${(n + 2)}^{2} - 4 (2 n) - 4 \cdot 1 = 0$

Step 1.1.2

Multiply $2$ by $- 4$ .

${(n + 2)}^{2} - 8 n - 4 \cdot 1 = 0$

Step 1.1.3

Multiply $- 4$ by $1$ .

${(n + 2)}^{2} - 8 n - 4 = 0$

Step 2

Use the quadratic formula to find the solutions.

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

Step 3

Substitute the values $a = 1$ , $b = - 4$ , and $c = 0$ into the quadratic formula and solve for $n$ .

$\frac{4 \pm \sqrt{{(- 4)}^{2} - 4 \cdot (1 \cdot 0)}}{2 \cdot 1}$

Step 4

Simplify.

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Step 4.1

Simplify the numerator.

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Step 4.1.1

Raise $- 4$ to the power of $2$ .

$n = \frac{4 \pm \sqrt{16 - 4 \cdot 1 \cdot 0}}{2 \cdot 1}$

Step 4.1.2

Multiply $- 4 \cdot 1 \cdot 0$ .

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Step 4.1.2.1

Multiply $- 4$ by $1$ .

$n = \frac{4 \pm \sqrt{16 - 4 \cdot 0}}{2 \cdot 1}$

Step 4.1.2.2

Multiply $- 4$ by $0$ .

$n = \frac{4 \pm \sqrt{16 + 0}}{2 \cdot 1}$

Step 4.1.3

Add $16$ and $0$ .

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

Step 4.1.4

Rewrite $16$ as $4^{2}$ .

$n = \frac{4 \pm \sqrt{4^{2}}}{2 \cdot 1}$

Step 4.1.5

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

$n = \frac{4 \pm 4}{2 \cdot 1}$

Step 4.2

Multiply $2$ by $1$ .

$n = \frac{4 \pm 4}{2}$

Step 4.3

Simplify $\frac{4 \pm 4}{2}$ .

$n = 2 \pm 2$

Step 5

The final answer is the combination of both solutions.

$n = 4, 0$