Solve 2x^2=(x+4)^2

$$2 x ^ { 2 } = ( x + 4 ) ^ { 2 }$$

Solve for x

$$x=4\sqrt{2}+4\approx 9.656854249$$

$$x=4-4\sqrt{2}\approx -1.656854249$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+4\right)^{2}.

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

Subtract x^{2} from both sides.

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

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

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

Subtract 8x from both sides.

$$x^{2}-8x=16$$

Subtract 16 from both sides.

$$x^{2}-8x-16=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -8 for b, and -16 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\left(-16\right)}}{2}$$

Square -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\left(-16\right)}}{2}$$

Multiply -4 times -16.

$$x=\frac{-\left(-8\right)±\sqrt{64+64}}{2}$$

Add 64 to 64.

$$x=\frac{-\left(-8\right)±\sqrt{128}}{2}$$

Take the square root of 128.

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

The opposite of -8 is 8.

$$x=\frac{8±8\sqrt{2}}{2}$$

Now solve the equation x=\frac{8±8\sqrt{2}}{2} when ± is plus. Add 8 to 8\sqrt{2}.

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

Divide 8+8\sqrt{2} by 2.

$$x=4\sqrt{2}+4$$

Now solve the equation x=\frac{8±8\sqrt{2}}{2} when ± is minus. Subtract 8\sqrt{2} from 8.

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

Divide 8-8\sqrt{2} by 2.

$$x=4-4\sqrt{2}$$

The equation is now solved.

$$x=4\sqrt{2}+4 x=4-4\sqrt{2}$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+4\right)^{2}.

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

Subtract x^{2} from both sides.

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

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

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

Subtract 8x from both sides.

$$x^{2}-8x=16$$

Divide -8, the coefficient of the x term, by 2 to get -4. Then add the square of -4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-8x+\left(-4\right)^{2}=16+\left(-4\right)^{2}$$

Square -4.

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

Add 16 to 16.

$$x^{2}-8x+16=32$$

Factor x^{2}-8x+16. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-4\right)^{2}=32$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-4\right)^{2}}=\sqrt{32}$$

Simplify.

$$x-4=4\sqrt{2} x-4=-4\sqrt{2}$$

Add 4 to both sides of the equation.

$$x=4\sqrt{2}+4 x=4-4\sqrt{2}$$