Basic Math Examples

$a^{2} + 2 a b + b^{2} = 16$

Step 1

Subtract $16$ from both sides of the equation.

$a^{2} + 2 a b + b^{2} - 16 = 0$

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 $a = 1$ , $b = 2 b$ , and $c = b^{2} - 16$ into the quadratic formula and solve for $a$ .

$\frac{- 2 b \pm \sqrt{{(2 b)}^{2} - 4 \cdot (1 \cdot (b^{2} - 16))}}{2 \cdot 1}$

Step 4

Simplify.

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Step 4.1

Simplify the numerator.

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Step 4.1.1

Add parentheses.

$a = \frac{- 2 b \pm \sqrt{{(2 b)}^{2} - 4 \cdot (1 \cdot (b^{2} - 16))}}{2 \cdot 1}$

Step 4.1.2

Let $u = 1 \cdot (b^{2} - 16)$ . Substitute $u$ for all occurrences of $1 \cdot (b^{2} - 16)$ .

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Step 4.1.2.1

Apply the product rule to $2 b$ .

$a = \frac{- 2 b \pm \sqrt{2^{2} b^{2} - 4 \cdot u}}{2 \cdot 1}$

Step 4.1.2.2

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

$a = \frac{- 2 b \pm \sqrt{4 b^{2} - 4 u}}{2 \cdot 1}$

Step 4.1.3

Factor $4$ out of $4 b^{2} - 4 u$ .

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Step 4.1.3.1

Factor $4$ out of $4 b^{2}$ .

$a = \frac{- 2 b \pm \sqrt{4 (b^{2}) - 4 u}}{2 \cdot 1}$

Step 4.1.3.2

Factor $4$ out of $- 4 u$ .

$a = \frac{- 2 b \pm \sqrt{4 (b^{2}) + 4 (- u)}}{2 \cdot 1}$

Step 4.1.3.3

Factor $4$ out of $4 (b^{2}) + 4 (- u)$ .

$a = \frac{- 2 b \pm \sqrt{4 (b^{2} - u)}}{2 \cdot 1}$

Step 4.1.4

Replace all occurrences of $u$ with $1 \cdot (b^{2} - 16)$ .

$a = \frac{- 2 b \pm \sqrt{4 (b^{2} - (1 \cdot (b^{2} - 16)))}}{2 \cdot 1}$

Step 4.1.5

Simplify.

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Step 4.1.5.1

Simplify each term.

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Step 4.1.5.1.1

Multiply $b^{2} - 16$ by $1$ .

$a = \frac{- 2 b \pm \sqrt{4 (b^{2} - (b^{2} - 16))}}{2 \cdot 1}$

Step 4.1.5.1.2

Apply the distributive property.

$a = \frac{- 2 b \pm \sqrt{4 (b^{2} - b^{2} + 16)}}{2 \cdot 1}$

Step 4.1.5.1.3

Multiply $- 1$ by $- 16$ .

$a = \frac{- 2 b \pm \sqrt{4 (b^{2} - b^{2} + 16)}}{2 \cdot 1}$

Step 4.1.5.2

Subtract $b^{2}$ from $b^{2}$ .

$a = \frac{- 2 b \pm \sqrt{4 (0 + 16)}}{2 \cdot 1}$

Step 4.1.5.3

Add $0$ and $16$ .

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

Step 4.1.6

Multiply $4$ by $16$ .

$a = \frac{- 2 b \pm \sqrt{64}}{2 \cdot 1}$

Step 4.1.7

Rewrite $64$ as $8^{2}$ .

$a = \frac{- 2 b \pm \sqrt{8^{2}}}{2 \cdot 1}$

Step 4.1.8

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

$a = \frac{- 2 b \pm 8}{2 \cdot 1}$

Step 4.2

Multiply $2$ by $1$ .

$a = \frac{- 2 b \pm 8}{2}$

Step 4.3

Simplify $\frac{- 2 b \pm 8}{2}$ .

$a = - b \pm 4$

Step 5

The final answer is the combination of both solutions.

$a = - b + 4$

$a = - b - 4$