Write as a Function of k 2x^2-kx+12=0

$2 x^{2} - k x + 12 = 0$

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 $a = 2$ , $b = - k$ , and $c = 12$ into the quadratic formula and solve for $x$ .

$\frac{k \pm \sqrt{{(- k)}^{2} - 4 \cdot (2 \cdot 12)}}{2 \cdot 2}$

Step 3

Simplify.

Simplify the numerator.

Apply the product rule to $- k$ .

$x = \frac{k \pm \sqrt{{(- 1)}^{2} k^{2} - 4 \cdot 2 \cdot 12}}{2 \cdot 2}$

Raise $- 1$ to the power of $2$ .

$x = \frac{k \pm \sqrt{1 k^{2} - 4 \cdot 2 \cdot 12}}{2 \cdot 2}$

Multiply $k^{2}$ by $1$ .

$x = \frac{k \pm \sqrt{k^{2} - 4 \cdot 2 \cdot 12}}{2 \cdot 2}$

Multiply $- 4 \cdot 2 \cdot 12$ .

Multiply $- 4$ by $2$ .

$x = \frac{k \pm \sqrt{k^{2} - 8 \cdot 12}}{2 \cdot 2}$

Multiply $- 8$ by $12$ .

$x = \frac{k \pm \sqrt{k^{2} - 96}}{2 \cdot 2}$

Multiply $2$ by $2$ .

$x = \frac{k \pm \sqrt{k^{2} - 96}}{4}$

Step 4

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

Simplify the numerator.

Apply the product rule to $- k$ .

$x = \frac{k \pm \sqrt{{(- 1)}^{2} k^{2} - 4 \cdot 2 \cdot 12}}{2 \cdot 2}$

Raise $- 1$ to the power of $2$ .

$x = \frac{k \pm \sqrt{1 k^{2} - 4 \cdot 2 \cdot 12}}{2 \cdot 2}$

Multiply $k^{2}$ by $1$ .

$x = \frac{k \pm \sqrt{k^{2} - 4 \cdot 2 \cdot 12}}{2 \cdot 2}$

Multiply $- 4 \cdot 2 \cdot 12$ .

Multiply $- 4$ by $2$ .

$x = \frac{k \pm \sqrt{k^{2} - 8 \cdot 12}}{2 \cdot 2}$

Multiply $- 8$ by $12$ .

$x = \frac{k \pm \sqrt{k^{2} - 96}}{2 \cdot 2}$

Multiply $2$ by $2$ .

$x = \frac{k \pm \sqrt{k^{2} - 96}}{4}$

Change the $\pm$ to $+$ .

$x = \frac{k + \sqrt{k^{2} - 96}}{4}$

Step 5

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

Simplify the numerator.

Apply the product rule to $- k$ .

$x = \frac{k \pm \sqrt{{(- 1)}^{2} k^{2} - 4 \cdot 2 \cdot 12}}{2 \cdot 2}$

Raise $- 1$ to the power of $2$ .

$x = \frac{k \pm \sqrt{1 k^{2} - 4 \cdot 2 \cdot 12}}{2 \cdot 2}$

Multiply $k^{2}$ by $1$ .

$x = \frac{k \pm \sqrt{k^{2} - 4 \cdot 2 \cdot 12}}{2 \cdot 2}$

Multiply $- 4 \cdot 2 \cdot 12$ .

Multiply $- 4$ by $2$ .

$x = \frac{k \pm \sqrt{k^{2} - 8 \cdot 12}}{2 \cdot 2}$

Multiply $- 8$ by $12$ .

$x = \frac{k \pm \sqrt{k^{2} - 96}}{2 \cdot 2}$

Multiply $2$ by $2$ .

$x = \frac{k \pm \sqrt{k^{2} - 96}}{4}$

Change the $\pm$ to $-$ .

$x = \frac{k - \sqrt{k^{2} - 96}}{4}$

Step 6

The final answer is the combination of both solutions.

$x = \frac{k + \sqrt{k^{2} - 96}}{4}$

$x = \frac{k - \sqrt{k^{2} - 96}}{4}$

Write as a Function of k 2x^2-kx+12=0

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