Write as a Function of b 2p^2+bp+32=0

$2 p^{2} + b p + 32 = 0$

Step 1

Use the quadratic formula to find the solutions.

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

Step 2

Substitute the values $a = 2$ , $b = b$ , and $c = 32$ into the quadratic formula and solve for $p$ .

$\frac{- b \pm \sqrt{b^{2} - 4 \cdot (2 \cdot 32)}}{2 \cdot 2}$

Step 3

Simplify.

Simplify the numerator.

Multiply $- 4 \cdot 2 \cdot 32$ .

Multiply $- 4$ by $2$ .

$p = \frac{- b \pm \sqrt{b^{2} - 8 \cdot 32}}{2 \cdot 2}$

Multiply $- 8$ by $32$ .

$p = \frac{- b \pm \sqrt{b^{2} - 256}}{2 \cdot 2}$

Rewrite $b^{2} - 256$ in a factored form.

Rewrite $256$ as $16^{2}$ .

$p = \frac{- b \pm \sqrt{b^{2} - 16^{2}}}{2 \cdot 2}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = b$ and $b = 16$ .

$p = \frac{- b \pm \sqrt{(b + 16) (b - 16)}}{2 \cdot 2}$

Multiply $2$ by $2$ .

$p = \frac{- b \pm \sqrt{(b + 16) (b - 16)}}{4}$

Step 4

Simplify the expression to solve for the $+$ portion of the $\pm$ .

Simplify the numerator.

Multiply $- 4 \cdot 2 \cdot 32$ .

Multiply $- 4$ by $2$ .

$p = \frac{- b \pm \sqrt{b^{2} - 8 \cdot 32}}{2 \cdot 2}$

Multiply $- 8$ by $32$ .

$p = \frac{- b \pm \sqrt{b^{2} - 256}}{2 \cdot 2}$

Rewrite $b^{2} - 256$ in a factored form.

Rewrite $256$ as $16^{2}$ .

$p = \frac{- b \pm \sqrt{b^{2} - 16^{2}}}{2 \cdot 2}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = b$ and $b = 16$ .

$p = \frac{- b \pm \sqrt{(b + 16) (b - 16)}}{2 \cdot 2}$

Multiply $2$ by $2$ .

$p = \frac{- b \pm \sqrt{(b + 16) (b - 16)}}{4}$

Change the $\pm$ to $+$ .

$p = \frac{- b + \sqrt{(b + 16) (b - 16)}}{4}$

Factor $- 1$ out of $- b$ .

$p = \frac{- (b) + \sqrt{(b + 16) (b - 16)}}{4}$

Factor $- 1$ out of $\sqrt{(b + 16) (b - 16)}$ .

$p = \frac{- (b) - 1 (- \sqrt{(b + 16) (b - 16)})}{4}$

Factor $- 1$ out of $- (b) - 1 (- \sqrt{(b + 16) (b - 16)})$ .

$p = \frac{- (b - \sqrt{(b + 16) (b - 16)})}{4}$

Rewrite $- (b - \sqrt{(b + 16) (b - 16)})$ as $- 1 (b - \sqrt{(b + 16) (b - 16)})$ .

$p = \frac{- 1 (b - \sqrt{(b + 16) (b - 16)})}{4}$

Move the negative in front of the fraction.

$p = - \frac{b - \sqrt{(b + 16) (b - 16)}}{4}$

Step 5

Simplify the expression to solve for the $-$ portion of the $\pm$ .

Simplify the numerator.

Multiply $- 4 \cdot 2 \cdot 32$ .

Multiply $- 4$ by $2$ .

$p = \frac{- b \pm \sqrt{b^{2} - 8 \cdot 32}}{2 \cdot 2}$

Multiply $- 8$ by $32$ .

$p = \frac{- b \pm \sqrt{b^{2} - 256}}{2 \cdot 2}$

Rewrite $b^{2} - 256$ in a factored form.

Rewrite $256$ as $16^{2}$ .

$p = \frac{- b \pm \sqrt{b^{2} - 16^{2}}}{2 \cdot 2}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = b$ and $b = 16$ .

$p = \frac{- b \pm \sqrt{(b + 16) (b - 16)}}{2 \cdot 2}$

Multiply $2$ by $2$ .

$p = \frac{- b \pm \sqrt{(b + 16) (b - 16)}}{4}$

Change the $\pm$ to $-$ .

$p = \frac{- b - \sqrt{(b + 16) (b - 16)}}{4}$

Factor $- 1$ out of $- b$ .

$p = \frac{- (b) - \sqrt{(b + 16) (b - 16)}}{4}$

Factor $- 1$ out of $- \sqrt{(b + 16) (b - 16)}$ .

$p = \frac{- (b) - (\sqrt{(b + 16) (b - 16)})}{4}$

Factor $- 1$ out of $- (b) - (\sqrt{(b + 16) (b - 16)})$ .

$p = \frac{- (b + \sqrt{(b + 16) (b - 16)})}{4}$

Rewrite $- (b + \sqrt{(b + 16) (b - 16)})$ as $- 1 (b + \sqrt{(b + 16) (b - 16)})$ .

$p = \frac{- 1 (b + \sqrt{(b + 16) (b - 16)})}{4}$

Move the negative in front of the fraction.

$p = - \frac{b + \sqrt{(b + 16) (b - 16)}}{4}$

Step 6

The final answer is the combination of both solutions.

$p = - \frac{b - \sqrt{(b + 16) (b - 16)}}{4}$

$p = - \frac{b + \sqrt{(b + 16) (b - 16)}}{4}$

Write as a Function of b 2p^2+bp+32=0

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