Solve for x |x^2-10x|=25

$| x^{2} - 10 x | = 25$

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

$x^{2} - 10 x = \pm 25$

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.

$x^{2} - 10 x = 25$

Subtract $25$ from both sides of the equation.

$x^{2} - 10 x - 25 = 0$

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 1$ , $b = - 10$ , and $c = - 25$ into the quadratic formula and solve for $x$ .

$\frac{10 \pm \sqrt{{(- 10)}^{2} - 4 \cdot (1 \cdot - 25)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

Raise $- 10$ to the power of $2$ .

$x = \frac{10 \pm \sqrt{100 - 4 \cdot 1 \cdot - 25}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$x = \frac{10 \pm \sqrt{100 - 4 \cdot - 25}}{2 \cdot 1}$

Multiply $- 4$ by $- 25$ .

$x = \frac{10 \pm \sqrt{100 + 100}}{2 \cdot 1}$

Add $100$ and $100$ .

$x = \frac{10 \pm \sqrt{200}}{2 \cdot 1}$

Rewrite $200$ as $10^{2} \cdot 2$ .

Factor $100$ out of $200$ .

$x = \frac{10 \pm \sqrt{100 (2)}}{2 \cdot 1}$

Rewrite $100$ as $10^{2}$ .

$x = \frac{10 \pm \sqrt{10^{2} \cdot 2}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{10 \pm 10 \sqrt{2}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{10 \pm 10 \sqrt{2}}{2}$

Simplify $\frac{10 \pm 10 \sqrt{2}}{2}$ .

$x = 5 \pm 5 \sqrt{2}$

The final answer is the combination of both solutions.

$x = 5 + 5 \sqrt{2}, 5 - 5 \sqrt{2}$

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

$x^{2} - 10 x = - 25$

Add $25$ to both sides of the equation.

$x^{2} - 10 x + 25 = 0$

Factor using the perfect square rule.

Rewrite $25$ as $5^{2}$ .

$x^{2} - 10 x + 5^{2} = 0$

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

$10 x = 2 \cdot x \cdot 5$

Rewrite the polynomial.

$x^{2} - 2 \cdot x \cdot 5 + 5^{2} = 0$

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 5$ .

${(x - 5)}^{2} = 0$

Set the $x - 5$ equal to $0$ .

$x - 5 = 0$

Add $5$ to both sides of the equation.

$x = 5$

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

$x = 5 + 5 \sqrt{2}, 5 - 5 \sqrt{2}, 5$

The result can be shown in multiple forms.

Exact Form:

$x = 5 + 5 \sqrt{2}, 5 - 5 \sqrt{2}, 5$

Decimal Form:

$x = 12.07106781 \dots, - 2.07106781 \dots, 5$

Solve for x |x^2-10x|=25

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