Solve x=sqrt{2x^2-2x-8}

$$x = \sqrt { 2 x ^ { 2 } - 2 x - 8 }$$

Solve for x

$$x=4$$

Solution Steps

Square both sides of the equation.

$$x^{2}=\left(\sqrt{2x^{2}-2x-8}\right)^{2}$$

Calculate \sqrt{2x^{2}-2x-8} to the power of 2 and get 2x^{2}-2x-8.

$$x^{2}=2x^{2}-2x-8$$

Subtract 2x^{2} from both sides.

$$x^{2}-2x^{2}=-2x-8$$

Combine x^{2} and -2x^{2} to get -x^{2}.

$$-x^{2}=-2x-8$$

Add 2x to both sides.

$$-x^{2}+2x=-8$$

Add 8 to both sides.

$$-x^{2}+2x+8=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+8. To find a and b, set up a system to be solved.

$$a+b=2 ab=-8=-8$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -8.

$$-1,8 -2,4$$

Calculate the sum for each pair.

$$-1+8=7 -2+4=2$$

The solution is the pair that gives sum 2.

$$a=4 b=-2$$

Rewrite -x^{2}+2x+8 as \left(-x^{2}+4x\right)+\left(-2x+8\right).

$$\left(-x^{2}+4x\right)+\left(-2x+8\right)$$

Factor out -x in the first and -2 in the second group.

$$-x\left(x-4\right)-2\left(x-4\right)$$

Factor out common term x-4 by using distributive property.

$$\left(x-4\right)\left(-x-2\right)$$

To find equation solutions, solve x-4=0 and -x-2=0.

$$x=4 x=-2$$

Substitute 4 for x in the equation x=\sqrt{2x^{2}-2x-8}.

$$4=\sqrt{2\times 4^{2}-2\times 4-8}$$

Simplify. The value x=4 satisfies the equation.

$$4=4$$

Substitute -2 for x in the equation x=\sqrt{2x^{2}-2x-8}.

$$-2=\sqrt{2\left(-2\right)^{2}-2\left(-2\right)-8}$$

Simplify. The value x=-2 does not satisfy the equation because the left and the right hand side have opposite signs.

$$-2=2$$

Equation x=\sqrt{2x^{2}-2x-8} has a unique solution.

$$x=4$$