Solve Using the Quadratic Formula k^2+6=6

$k^{2} + 6 = 6$

Step 1

Move all terms to the left side of the equation and simplify.

Subtract $6$ from both sides of the equation.

$k^{2} + 6 - 6 = 0$

Combine the opposite terms in $k^{2} + 6 - 6$ .

Subtract $6$ from $6$ .

$k^{2} + 0 = 0$

Add $k^{2}$ and $0$ .

$k^{2} = 0$

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 $a = 1$ , $b = 0$ , and $c = 0$ into the quadratic formula and solve for $k$ .

$\frac{0 \pm \sqrt{0^{2} - 4 \cdot (1 \cdot 0)}}{2 \cdot 1}$

Step 4

Simplify.

Simplify the numerator.

Raising $0$ to any positive power yields $0$ .

$k = \frac{0 \pm \sqrt{0 - 4 \cdot 1 \cdot 0}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$k = \frac{0 \pm \sqrt{0 - 4 \cdot 0}}{2 \cdot 1}$

Multiply $- 4$ by $0$ .

$k = \frac{0 \pm \sqrt{0 + 0}}{2 \cdot 1}$

Add $0$ and $0$ .

$k = \frac{0 \pm \sqrt{0}}{2 \cdot 1}$

Rewrite $0$ as $0^{2}$ .

$k = \frac{0 \pm \sqrt{0^{2}}}{2 \cdot 1}$

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

$k = \frac{0 \pm 0}{2 \cdot 1}$

Plus or minus $0$ is $0$ .

$k = \frac{0}{2 \cdot 1}$

Multiply $2$ by $1$ .

$k = \frac{0}{2}$

Divide $0$ by $2$ .

$k = 0$

Step 5

The final answer is the combination of both solutions.

$k = 0$ Double roots