Solve 2x^2=(8-x)^2

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

Solve for x

$$x=8\sqrt{2}-8\approx 3.313708499$$

$$x=-8\sqrt{2}-8\approx -19.313708499$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(8-x\right)^{2}.

$$2x^{2}=64-16x+x^{2}$$

Subtract 64 from both sides.

$$2x^{2}-64=-16x+x^{2}$$

Add 16x to both sides.

$$2x^{2}-64+16x=x^{2}$$

Subtract x^{2} from both sides.

$$2x^{2}-64+16x-x^{2}=0$$

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

$$x^{2}-64+16x=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$x^{2}+16x-64=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 16 for b, and -64 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-16±\sqrt{16^{2}-4\left(-64\right)}}{2}$$

Square 16.

$$x=\frac{-16±\sqrt{256-4\left(-64\right)}}{2}$$

Multiply -4 times -64.

$$x=\frac{-16±\sqrt{256+256}}{2}$$

Add 256 to 256.

$$x=\frac{-16±\sqrt{512}}{2}$$

Take the square root of 512.

$$x=\frac{-16±16\sqrt{2}}{2}$$

Now solve the equation x=\frac{-16±16\sqrt{2}}{2} when ± is plus. Add -16 to 16\sqrt{2}.

$$x=\frac{16\sqrt{2}-16}{2}$$

Divide -16+16\sqrt{2} by 2.

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

Now solve the equation x=\frac{-16±16\sqrt{2}}{2} when ± is minus. Subtract 16\sqrt{2} from -16.

$$x=\frac{-16\sqrt{2}-16}{2}$$

Divide -16-16\sqrt{2} by 2.

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

The equation is now solved.

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

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(8-x\right)^{2}.

$$2x^{2}=64-16x+x^{2}$$

Add 16x to both sides.

$$2x^{2}+16x=64+x^{2}$$

Subtract x^{2} from both sides.

$$2x^{2}+16x-x^{2}=64$$

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

$$x^{2}+16x=64$$

Divide 16, the coefficient of the x term, by 2 to get 8. Then add the square of 8 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+16x+8^{2}=64+8^{2}$$

Square 8.

$$x^{2}+16x+64=64+64$$

Add 64 to 64.

$$x^{2}+16x+64=128$$

Factor x^{2}+16x+64. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+8\right)^{2}=128$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+8\right)^{2}}=\sqrt{128}$$

Simplify.

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

Subtract 8 from both sides of the equation.

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