Finite Math Examples

$f (x) = 4 x^{2} - 1 + 2 x$

Step 1

Set $4 x^{2} - 1 + 2 x$ equal to $0$ .

$4 x^{2} - 1 + 2 x = 0$

Step 2

Solve for $x$ .

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Step 2.1

Use the quadratic formula to find the solutions.

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

Step 2.2

Substitute the values $a = 4$ , $b = 2$ , and $c = - 1$ into the quadratic formula and solve for $x$ .

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

Step 2.3

Simplify.

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Step 2.3.1

Simplify the numerator.

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Step 2.3.1.1

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

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

Step 2.3.1.2

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

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Step 2.3.1.2.1

Multiply $- 4$ by $4$ .

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

Step 2.3.1.2.2

Multiply $- 16$ by $- 1$ .

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

Step 2.3.1.3

Add $4$ and $16$ .

$x = \frac{- 2 \pm \sqrt{20}}{2 \cdot 4}$

Step 2.3.1.4

Rewrite $20$ as $2^{2} \cdot 5$ .

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Step 2.3.1.4.1

Factor $4$ out of $20$ .

$x = \frac{- 2 \pm \sqrt{4 (5)}}{2 \cdot 4}$

Step 2.3.1.4.2

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

$x = \frac{- 2 \pm \sqrt{2^{2} \cdot 5}}{2 \cdot 4}$

Step 2.3.1.5

Pull terms out from under the radical.

$x = \frac{- 2 \pm 2 \sqrt{5}}{2 \cdot 4}$

Step 2.3.2

Multiply $2$ by $4$ .

$x = \frac{- 2 \pm 2 \sqrt{5}}{8}$

Step 2.3.3

Simplify $\frac{- 2 \pm 2 \sqrt{5}}{8}$ .

$x = \frac{- 1 \pm \sqrt{5}}{4}$

Step 2.4

The final answer is the combination of both solutions.

$x = - \frac{1 - \sqrt{5}}{4}, - \frac{1 + \sqrt{5}}{4}$

$x = \frac{- 1 \pm \sqrt{5}}{4}$

Step 3

The result can be shown in multiple forms.

Exact Form:

$x = \frac{- 1 \pm \sqrt{5}}{4}$

Decimal Form:

$x = 0.30901699 \dots, - 0.80901699 \dots$

Step 4