Solve for x |x^2-6x|=9

$| x^{2} - 6 x | = 9$

Remove the absolute value term. This creates a $\pm$ on the right side of the equation because $| x | = \pm x$ .

$x^{2} - 6 x = \pm 9$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$x^{2} - 6 x = 9$

Subtract $9$ from both sides of the equation.

$x^{2} - 6 x - 9 = 0$

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 1$ , $b = - 6$ , and $c = - 9$ into the quadratic formula and solve for $x$ .

$\frac{6 \pm \sqrt{{(- 6)}^{2} - 4 \cdot (1 \cdot - 9)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

Raise $- 6$ to the power of $2$ .

$x = \frac{6 \pm \sqrt{36 - 4 \cdot 1 \cdot - 9}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$x = \frac{6 \pm \sqrt{36 - 4 \cdot - 9}}{2 \cdot 1}$

Multiply $- 4$ by $- 9$ .

$x = \frac{6 \pm \sqrt{36 + 36}}{2 \cdot 1}$

Add $36$ and $36$ .

$x = \frac{6 \pm \sqrt{72}}{2 \cdot 1}$

Rewrite $72$ as $6^{2} \cdot 2$ .

Factor $36$ out of $72$ .

$x = \frac{6 \pm \sqrt{36 (2)}}{2 \cdot 1}$

Rewrite $36$ as $6^{2}$ .

$x = \frac{6 \pm \sqrt{6^{2} \cdot 2}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{6 \pm 6 \sqrt{2}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{6 \pm 6 \sqrt{2}}{2}$

Simplify $\frac{6 \pm 6 \sqrt{2}}{2}$ .

$x = 3 \pm 3 \sqrt{2}$

The final answer is the combination of both solutions.

$x = 3 + 3 \sqrt{2}, 3 - 3 \sqrt{2}$

Next, use the negative value of the $\pm$ to find the second solution.

$x^{2} - 6 x = - 9$

Add $9$ to both sides of the equation.

$x^{2} - 6 x + 9 = 0$

Factor using the perfect square rule.

Rewrite $9$ as $3^{2}$ .

$x^{2} - 6 x + 3^{2} = 0$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$6 x = 2 \cdot x \cdot 3$

Rewrite the polynomial.

$x^{2} - 2 \cdot x \cdot 3 + 3^{2} = 0$

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 3$ .

${(x - 3)}^{2} = 0$

Set the $x - 3$ equal to $0$ .

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

The complete solution is the result of both the positive and negative portions of the solution.

$x = 3 + 3 \sqrt{2}, 3 - 3 \sqrt{2}, 3$

The result can be shown in multiple forms.

Exact Form:

$x = 3 + 3 \sqrt{2}, 3 - 3 \sqrt{2}, 3$

Decimal Form:

$x = 7.24264068 \dots, - 1.24264068 \dots, 3$

Solve for x |x^2-6x|=9

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