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

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

Solve for x

$$x=2\sqrt{3}+4\approx 7.464101615$$

$$x=4-2\sqrt{3}\approx 0.535898385$$

Steps Using the Quadratic Formula

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

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

To find the opposite of 4-8x+4x^{2}, find the opposite of each term.

$$4x^{2}-4+8x-4x^{2}=x^{2}$$

Combine 4x^{2} and -4x^{2} to get 0.

$$-4+8x=x^{2}$$

Subtract x^{2} from both sides.

$$-4+8x-x^{2}=0$$

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.

$$-x^{2}+8x-4=0$$

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

$$x=\frac{-8±\sqrt{8^{2}-4\left(-1\right)\left(-4\right)}}{2\left(-1\right)}$$

Square 8.

$$x=\frac{-8±\sqrt{64-4\left(-1\right)\left(-4\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-8±\sqrt{64+4\left(-4\right)}}{2\left(-1\right)}$$

Multiply 4 times -4.

$$x=\frac{-8±\sqrt{64-16}}{2\left(-1\right)}$$

Add 64 to -16.

$$x=\frac{-8±\sqrt{48}}{2\left(-1\right)}$$

Take the square root of 48.

$$x=\frac{-8±4\sqrt{3}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-8±4\sqrt{3}}{-2}$$

Now solve the equation x=\frac{-8±4\sqrt{3}}{-2} when ± is plus. Add -8 to 4\sqrt{3}.

$$x=\frac{4\sqrt{3}-8}{-2}$$

Divide -8+4\sqrt{3} by -2.

$$x=4-2\sqrt{3}$$

Now solve the equation x=\frac{-8±4\sqrt{3}}{-2} when ± is minus. Subtract 4\sqrt{3} from -8.

$$x=\frac{-4\sqrt{3}-8}{-2}$$

Divide -8-4\sqrt{3} by -2.

$$x=2\sqrt{3}+4$$

The equation is now solved.

$$x=4-2\sqrt{3} x=2\sqrt{3}+4$$

Steps for Completing the Square

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

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

To find the opposite of 4-8x+4x^{2}, find the opposite of each term.

$$4x^{2}-4+8x-4x^{2}=x^{2}$$

Combine 4x^{2} and -4x^{2} to get 0.

$$-4+8x=x^{2}$$

Subtract x^{2} from both sides.

$$-4+8x-x^{2}=0$$

Add 4 to both sides. Anything plus zero gives itself.

$$8x-x^{2}=4$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-x^{2}+8x=4$$

Divide both sides by -1.

$$\frac{-x^{2}+8x}{-1}=\frac{4}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{8}{-1}x=\frac{4}{-1}$$

Divide 8 by -1.

$$x^{2}-8x=\frac{4}{-1}$$

Divide 4 by -1.

$$x^{2}-8x=-4$$

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.

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

Square -4.

$$x^{2}-8x+16=-4+16$$

Add -4 to 16.

$$x^{2}-8x+16=12$$

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}.

$$\left(x-4\right)^{2}=12$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-4\right)^{2}}=\sqrt{12}$$

Simplify.

$$x-4=2\sqrt{3} x-4=-2\sqrt{3}$$

Add 4 to both sides of the equation.

$$x=2\sqrt{3}+4 x=4-2\sqrt{3}$$