Write as a Function of b 4p^2+bp+4=0

$4 p^{2} + b p + 4 = 0$

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 4$ , $b = b$ , and $c = 4$ into the quadratic formula and solve for $p$ .

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

Simplify.

Simplify the numerator.

Rewrite $4 \cdot 4 \cdot 4$ as ${(2 \cdot 2 \cdot 2)}^{2}$ .

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

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 $b = 2 \cdot 2 \cdot 2$ .

$p = \frac{- b \pm \sqrt{(b + 2 \cdot 2 \cdot 2) (b - (2 \cdot 2 \cdot 2))}}{2 \cdot 4}$

Simplify.

Multiply $2$ by $2$ .

$p = \frac{- b \pm \sqrt{(b + 4 \cdot 2) (b - (2 \cdot 2 \cdot 2))}}{2 \cdot 4}$

Multiply $4$ by $2$ .

$p = \frac{- b \pm \sqrt{(b + 8) (b - (2 \cdot 2 \cdot 2))}}{2 \cdot 4}$

Multiply $2$ by $2$ .

$p = \frac{- b \pm \sqrt{(b + 8) (b - (4 \cdot 2))}}{2 \cdot 4}$

Multiply $4$ by $2$ .

$p = \frac{- b \pm \sqrt{(b + 8) (b - 1 \cdot 8)}}{2 \cdot 4}$

Multiply $- 1$ by $8$ .

$p = \frac{- b \pm \sqrt{(b + 8) (b - 8)}}{2 \cdot 4}$

Multiply $2$ by $4$ .

$p = \frac{- b \pm \sqrt{(b + 8) (b - 8)}}{8}$

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

Simplify the numerator.

Rewrite $4 \cdot 4 \cdot 4$ as ${(2 \cdot 2 \cdot 2)}^{2}$ .

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

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 $b = 2 \cdot 2 \cdot 2$ .

$p = \frac{- b \pm \sqrt{(b + 2 \cdot 2 \cdot 2) (b - (2 \cdot 2 \cdot 2))}}{2 \cdot 4}$

Simplify.

Multiply $2$ by $2$ .

$p = \frac{- b \pm \sqrt{(b + 4 \cdot 2) (b - (2 \cdot 2 \cdot 2))}}{2 \cdot 4}$

Multiply $4$ by $2$ .

$p = \frac{- b \pm \sqrt{(b + 8) (b - (2 \cdot 2 \cdot 2))}}{2 \cdot 4}$

Multiply $2$ by $2$ .

$p = \frac{- b \pm \sqrt{(b + 8) (b - (4 \cdot 2))}}{2 \cdot 4}$

Multiply $4$ by $2$ .

$p = \frac{- b \pm \sqrt{(b + 8) (b - 1 \cdot 8)}}{2 \cdot 4}$

Multiply $- 1$ by $8$ .

$p = \frac{- b \pm \sqrt{(b + 8) (b - 8)}}{2 \cdot 4}$

Multiply $2$ by $4$ .

$p = \frac{- b \pm \sqrt{(b + 8) (b - 8)}}{8}$

Change the $\pm$ to $+$ .

$p = \frac{- b + \sqrt{(b + 8) (b - 8)}}{8}$

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

$p = \frac{- (b) + \sqrt{(b + 8) (b - 8)}}{8}$

Factor $- 1$ out of $\sqrt{(b + 8) (b - 8)}$ .

$p = \frac{- (b) - 1 (- \sqrt{(b + 8) (b - 8)})}{8}$

Factor $- 1$ out of $- (b) - 1 (- \sqrt{(b + 8) (b - 8)})$ .

$p = \frac{- (b - \sqrt{(b + 8) (b - 8)})}{8}$

Rewrite $- (b - \sqrt{(b + 8) (b - 8)})$ as $- 1 (b - \sqrt{(b + 8) (b - 8)})$ .

$p = \frac{- 1 (b - \sqrt{(b + 8) (b - 8)})}{8}$

Move the negative in front of the fraction.

$p = - \frac{b - \sqrt{(b + 8) (b - 8)}}{8}$

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

Simplify the numerator.

Rewrite $4 \cdot 4 \cdot 4$ as ${(2 \cdot 2 \cdot 2)}^{2}$ .

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

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 $b = 2 \cdot 2 \cdot 2$ .

$p = \frac{- b \pm \sqrt{(b + 2 \cdot 2 \cdot 2) (b - (2 \cdot 2 \cdot 2))}}{2 \cdot 4}$

Simplify.

Multiply $2$ by $2$ .

$p = \frac{- b \pm \sqrt{(b + 4 \cdot 2) (b - (2 \cdot 2 \cdot 2))}}{2 \cdot 4}$

Multiply $4$ by $2$ .

$p = \frac{- b \pm \sqrt{(b + 8) (b - (2 \cdot 2 \cdot 2))}}{2 \cdot 4}$

Multiply $2$ by $2$ .

$p = \frac{- b \pm \sqrt{(b + 8) (b - (4 \cdot 2))}}{2 \cdot 4}$

Multiply $4$ by $2$ .

$p = \frac{- b \pm \sqrt{(b + 8) (b - 1 \cdot 8)}}{2 \cdot 4}$

Multiply $- 1$ by $8$ .

$p = \frac{- b \pm \sqrt{(b + 8) (b - 8)}}{2 \cdot 4}$

Multiply $2$ by $4$ .

$p = \frac{- b \pm \sqrt{(b + 8) (b - 8)}}{8}$

Change the $\pm$ to $-$ .

$p = \frac{- b - \sqrt{(b + 8) (b - 8)}}{8}$

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

$p = \frac{- (b) - \sqrt{(b + 8) (b - 8)}}{8}$

Factor $- 1$ out of $- \sqrt{(b + 8) (b - 8)}$ .

$p = \frac{- (b) - (\sqrt{(b + 8) (b - 8)})}{8}$

Factor $- 1$ out of $- (b) - (\sqrt{(b + 8) (b - 8)})$ .

$p = \frac{- (b + \sqrt{(b + 8) (b - 8)})}{8}$

Rewrite $- (b + \sqrt{(b + 8) (b - 8)})$ as $- 1 (b + \sqrt{(b + 8) (b - 8)})$ .

$p = \frac{- 1 (b + \sqrt{(b + 8) (b - 8)})}{8}$

Move the negative in front of the fraction.

$p = - \frac{b + \sqrt{(b + 8) (b - 8)}}{8}$

The final answer is the combination of both solutions.

$p = - \frac{b - \sqrt{(b + 8) (b - 8)}}{8}$

$p = - \frac{b + \sqrt{(b + 8) (b - 8)}}{8}$

Write as a Function of b 4p^2+bp+4=0

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