Write as a Function of c x^2+4x+c=0

$x^{2} + 4 x + c = 0$

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 $a = 1$ , $b = 4$ , and $c = c$ into the quadratic formula and solve for $x$ .

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

Step 3

Simplify.

Simplify the numerator.

Raise $4$ to the power of $2$ .

$x = \frac{- 4 \pm \sqrt{16 - 4 \cdot 1 \cdot c}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$x = \frac{- 4 \pm \sqrt{16 - 4 \cdot c}}{2 \cdot 1}$

Factor $4$ out of $16 - 4 c$ .

Factor $4$ out of $16$ .

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

Factor $4$ out of $- 4 c$ .

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

Factor $4$ out of $4 (4) + 4 (- c)$ .

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

Rewrite $4 (4 - c)$ as $2^{2} (2^{2} - c)$ .

Rewrite $4$ as $2^{2}$ .

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

Rewrite $4$ as $2^{2}$ .

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

Pull terms out from under the radical.

$x = \frac{- 4 \pm 2 \sqrt{2^{2} - c}}{2 \cdot 1}$

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

$x = \frac{- 4 \pm 2 \sqrt{4 - c}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 4 \pm 2 \sqrt{4 - c}}{2}$

Simplify $\frac{- 4 \pm 2 \sqrt{4 - c}}{2}$ .

$x = - 2 \pm \sqrt{4 - c}$

Step 4

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

Simplify the numerator.

Raise $4$ to the power of $2$ .

$x = \frac{- 4 \pm \sqrt{16 - 4 \cdot 1 \cdot c}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$x = \frac{- 4 \pm \sqrt{16 - 4 \cdot c}}{2 \cdot 1}$

Factor $4$ out of $16 - 4 c$ .

Factor $4$ out of $16$ .

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

Factor $4$ out of $- 4 c$ .

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

Factor $4$ out of $4 (4) + 4 (- c)$ .

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

Rewrite $4 (4 - c)$ as $2^{2} (2^{2} - c)$ .

Rewrite $4$ as $2^{2}$ .

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

Rewrite $4$ as $2^{2}$ .

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

Pull terms out from under the radical.

$x = \frac{- 4 \pm 2 \sqrt{2^{2} - c}}{2 \cdot 1}$

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

$x = \frac{- 4 \pm 2 \sqrt{4 - c}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 4 \pm 2 \sqrt{4 - c}}{2}$

Simplify $\frac{- 4 \pm 2 \sqrt{4 - c}}{2}$ .

$x = - 2 \pm \sqrt{4 - c}$

Change the $\pm$ to $+$ .

$x = - 2 + \sqrt{4 - c}$

Step 5

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

Simplify the numerator.

Raise $4$ to the power of $2$ .

$x = \frac{- 4 \pm \sqrt{16 - 4 \cdot 1 \cdot c}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$x = \frac{- 4 \pm \sqrt{16 - 4 \cdot c}}{2 \cdot 1}$

Factor $4$ out of $16 - 4 c$ .

Factor $4$ out of $16$ .

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

Factor $4$ out of $- 4 c$ .

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

Factor $4$ out of $4 (4) + 4 (- c)$ .

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

Rewrite $4 (4 - c)$ as $2^{2} (2^{2} - c)$ .

Rewrite $4$ as $2^{2}$ .

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

Rewrite $4$ as $2^{2}$ .

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

Pull terms out from under the radical.

$x = \frac{- 4 \pm 2 \sqrt{2^{2} - c}}{2 \cdot 1}$

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

$x = \frac{- 4 \pm 2 \sqrt{4 - c}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 4 \pm 2 \sqrt{4 - c}}{2}$

Simplify $\frac{- 4 \pm 2 \sqrt{4 - c}}{2}$ .

$x = - 2 \pm \sqrt{4 - c}$

Change the $\pm$ to $-$ .

$x = - 2 - \sqrt{4 - c}$

Step 6

The final answer is the combination of both solutions.

$x = - 2 + \sqrt{4 - c}$

$x = - 2 - \sqrt{4 - c}$

Write as a Function of c x^2+4x+c=0

Related Questions

Support LCMGCF.com by disabling your adblocker.