Algebra Examples

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

Step 1

Since $x$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

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

Step 2

Rewrite using the commutative property of multiplication.

$4 x^{2} + 2 x - 4 = - f x$

Step 3

Add $f x$ to both sides of the equation.

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

Step 4

Use the quadratic formula to find the solutions.

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

Step 5

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

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

Step 6

Simplify.

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Step 6.1

Simplify the numerator.

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Step 6.1.1

Apply the distributive property.

$x = \frac{- 1 \cdot 2 - f \pm \sqrt{{(2 + f)}^{2} - 4 \cdot 4 \cdot - 4}}{2 \cdot 4}$

Step 6.1.2

Multiply $- 1$ by $2$ .

$x = \frac{- 2 - f \pm \sqrt{{(2 + f)}^{2} - 4 \cdot 4 \cdot - 4}}{2 \cdot 4}$

Step 6.1.3

Rewrite ${(2 + f)}^{2}$ as $(2 + f) (2 + f)$ .

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

Step 6.1.4

Expand $(2 + f) (2 + f)$ using the FOIL Method.

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Step 6.1.4.1

Apply the distributive property.

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

Step 6.1.4.2

Apply the distributive property.

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

Step 6.1.4.3

Apply the distributive property.

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

Step 6.1.5

Simplify and combine like terms.

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Step 6.1.5.1

Simplify each term.

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Step 6.1.5.1.1

Multiply $2$ by $2$ .

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

Step 6.1.5.1.2

Move $2$ to the left of $f$ .

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

Step 6.1.5.1.3

Multiply $f$ by $f$ .

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

Step 6.1.5.2

Add $2 f$ and $2 f$ .

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

Step 6.1.6

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

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Step 6.1.6.1

Multiply $- 4$ by $4$ .

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

Step 6.1.6.2

Multiply $- 16$ by $- 4$ .

$x = \frac{- 2 - f \pm \sqrt{4 + 4 f + f^{2} + 64}}{2 \cdot 4}$

Step 6.1.7

Add $4$ and $64$ .

$x = \frac{- 2 - f \pm \sqrt{4 f + f^{2} + 68}}{2 \cdot 4}$

Step 6.1.8

Reorder terms.

$x = \frac{- 2 - f \pm \sqrt{f^{2} + 4 f + 68}}{2 \cdot 4}$

Step 6.2

Multiply $2$ by $4$ .

$x = \frac{- 2 - f \pm \sqrt{f^{2} + 4 f + 68}}{8}$

Step 7

The final answer is the combination of both solutions.

$x = - \frac{2 + f - \sqrt{f^{2} + 4 f + 68}}{8}$

$x = - \frac{2 + f + \sqrt{f^{2} + 4 f + 68}}{8}$