Basic Math Examples

$f (f + g) = f + g$

Step 1

Simplify $f (f + g)$ .

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Step 1.1

Rewrite.

$0 + 0 + f (f + g) = f + g$

Step 1.2

Simplify by adding zeros.

$f (f + g) = f + g$

Step 1.3

Apply the distributive property.

$f \cdot f + f g = f + g$

Step 1.4

Multiply $f$ by $f$ .

$f^{2} + f g = f + g$

Step 2

Subtract $f$ from both sides of the equation.

$f^{2} + f g - f = g$

Step 3

Subtract $g$ from both sides of the equation.

$f^{2} + f g - f - g = 0$

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 $a = 1$ , $b = g - 1$ , and $c = - g$ into the quadratic formula and solve for $f$ .

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

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.

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

Step 6.1.2

Multiply $- 1$ by $- 1$ .

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

Step 6.1.3

Rewrite ${(g - 1)}^{2}$ as $(g - 1) (g - 1)$ .

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

Step 6.1.4

Expand $(g - 1) (g - 1)$ using the FOIL Method.

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Step 6.1.4.1

Apply the distributive property.

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

Step 6.1.4.2

Apply the distributive property.

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

Step 6.1.4.3

Apply the distributive property.

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

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 $g$ by $g$ .

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

Step 6.1.5.1.2

Move $- 1$ to the left of $g$ .

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

Step 6.1.5.1.3

Rewrite $- 1 g$ as $- g$ .

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

Step 6.1.5.1.4

Rewrite $- 1 g$ as $- g$ .

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

Step 6.1.5.1.5

Multiply $- 1$ by $- 1$ .

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

Step 6.1.5.2

Subtract $g$ from $- g$ .

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

Step 6.1.6

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

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Step 6.1.6.1

Multiply $- 4$ by $1$ .

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

Step 6.1.6.2

Multiply $- 4$ by $- 1$ .

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

Step 6.1.7

Add $- 2 g$ and $4 g$ .

$f = \frac{- g + 1 \pm \sqrt{g^{2} + 2 g + 1}}{2 \cdot 1}$

Step 6.1.8

Factor using the perfect square rule.

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Step 6.1.8.1

Rewrite $1$ as $1^{2}$ .

$f = \frac{- g + 1 \pm \sqrt{g^{2} + 2 g + 1^{2}}}{2 \cdot 1}$

Step 6.1.8.2

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$2 g = 2 \cdot g \cdot 1$

Step 6.1.8.3

Rewrite the polynomial.

$f = \frac{- g + 1 \pm \sqrt{g^{2} + 2 \cdot g \cdot 1 + 1^{2}}}{2 \cdot 1}$

Step 6.1.8.4

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = g$ and $b = 1$ .

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

Step 6.1.9

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

$f = \frac{- g + 1 \pm (g + 1)}{2 \cdot 1}$

Step 6.2

Multiply $2$ by $1$ .

$f = \frac{- g + 1 \pm (g + 1)}{2}$

Step 7

The final answer is the combination of both solutions.

$f = 1$

$f = - g$