Solve x=2(1-2x/4)^2

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

Solve for x

$$x=\sqrt{5}+3\approx 5.236067977$$

$$x=3-\sqrt{5}\approx 0.763932023$$

Steps Using the Quadratic Formula

Divide 2x by 4 to get \frac{1}{2}x.

$$x=2\left(1-\frac{1}{2}x\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(1-\frac{1}{2}x\right)^{2}.

$$x=2\left(1+2\left(-\frac{1}{2}x\right)+\left(-\frac{1}{2}x\right)^{2}\right)$$

Calculate -\frac{1}{2}x to the power of 2 and get \left(\frac{1}{2}x\right)^{2}.

$$x=2\left(1+2\left(-\frac{1}{2}x\right)+\left(\frac{1}{2}x\right)^{2}\right)$$

Expand \left(\frac{1}{2}x\right)^{2}.

$$x=2\left(1+2\left(-\frac{1}{2}x\right)+\left(\frac{1}{2}\right)^{2}x^{2}\right)$$

Calculate \frac{1}{2} to the power of 2 and get \frac{1}{4}.

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

Use the distributive property to multiply 2 by 1+2\left(-\frac{1}{2}x\right)+\frac{1}{4}x^{2}.

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

Subtract 2 from both sides.

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

Subtract 4\left(-\frac{1}{2}x\right) from both sides.

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

Subtract \frac{1}{2}x^{2} from both sides.

$$x-2-4\left(-\frac{1}{2}x\right)-\frac{1}{2}x^{2}=0$$

Multiply -1 and 4 to get -4.

$$x-2-4\left(-1\right)\times \frac{1}{2}x-\frac{1}{2}x^{2}=0$$

Multiply -4 and -1 to get 4.

$$x-2+4\times \frac{1}{2}x-\frac{1}{2}x^{2}=0$$

Multiply 4 and \frac{1}{2} to get 2.

$$x-2+2x-\frac{1}{2}x^{2}=0$$

Combine x and 2x to get 3x.

$$3x-2-\frac{1}{2}x^{2}=0$$

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.

$$-\frac{1}{2}x^{2}+3x-2=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -\frac{1}{2} for a, 3 for b, and -2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square 3.

$$x=\frac{-3±\sqrt{9-4\left(-\frac{1}{2}\right)\left(-2\right)}}{2\left(-\frac{1}{2}\right)}$$

Multiply -4 times -\frac{1}{2}.

$$x=\frac{-3±\sqrt{9+2\left(-2\right)}}{2\left(-\frac{1}{2}\right)}$$

Multiply 2 times -2.

$$x=\frac{-3±\sqrt{9-4}}{2\left(-\frac{1}{2}\right)}$$

Add 9 to -4.

$$x=\frac{-3±\sqrt{5}}{2\left(-\frac{1}{2}\right)}$$

Multiply 2 times -\frac{1}{2}.

$$x=\frac{-3±\sqrt{5}}{-1}$$

Now solve the equation x=\frac{-3±\sqrt{5}}{-1} when ± is plus. Add -3 to \sqrt{5}.

$$x=\frac{\sqrt{5}-3}{-1}$$

Divide -3+\sqrt{5} by -1.

$$x=3-\sqrt{5}$$

Now solve the equation x=\frac{-3±\sqrt{5}}{-1} when ± is minus. Subtract \sqrt{5} from -3.

$$x=\frac{-\sqrt{5}-3}{-1}$$

Divide -3-\sqrt{5} by -1.

$$x=\sqrt{5}+3$$

The equation is now solved.

$$x=3-\sqrt{5} x=\sqrt{5}+3$$

Steps for Completing the Square

Divide 2x by 4 to get \frac{1}{2}x.

$$x=2\left(1-\frac{1}{2}x\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(1-\frac{1}{2}x\right)^{2}.

$$x=2\left(1+2\left(-\frac{1}{2}x\right)+\left(-\frac{1}{2}x\right)^{2}\right)$$

Calculate -\frac{1}{2}x to the power of 2 and get \left(\frac{1}{2}x\right)^{2}.

$$x=2\left(1+2\left(-\frac{1}{2}x\right)+\left(\frac{1}{2}x\right)^{2}\right)$$

Expand \left(\frac{1}{2}x\right)^{2}.

$$x=2\left(1+2\left(-\frac{1}{2}x\right)+\left(\frac{1}{2}\right)^{2}x^{2}\right)$$

Calculate \frac{1}{2} to the power of 2 and get \frac{1}{4}.

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

Use the distributive property to multiply 2 by 1+2\left(-\frac{1}{2}x\right)+\frac{1}{4}x^{2}.

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

Subtract 4\left(-\frac{1}{2}x\right) from both sides.

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

Subtract \frac{1}{2}x^{2} from both sides.

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

Multiply -1 and 4 to get -4.

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

Multiply -4 and -1 to get 4.

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

Multiply 4 and \frac{1}{2} to get 2.

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

Combine x and 2x to get 3x.

$$3x-\frac{1}{2}x^{2}=2$$

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.

$$-\frac{1}{2}x^{2}+3x=2$$

Multiply both sides by -2.

$$\frac{-\frac{1}{2}x^{2}+3x}{-\frac{1}{2}}=\frac{2}{-\frac{1}{2}}$$

Dividing by -\frac{1}{2} undoes the multiplication by -\frac{1}{2}.

$$x^{2}+\frac{3}{-\frac{1}{2}}x=\frac{2}{-\frac{1}{2}}$$

Divide 3 by -\frac{1}{2} by multiplying 3 by the reciprocal of -\frac{1}{2}.

$$x^{2}-6x=\frac{2}{-\frac{1}{2}}$$

Divide 2 by -\frac{1}{2} by multiplying 2 by the reciprocal of -\frac{1}{2}.

$$x^{2}-6x=-4$$

Divide -6, the coefficient of the x term, by 2 to get -3. Then add the square of -3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-6x+\left(-3\right)^{2}=-4+\left(-3\right)^{2}$$

Square -3.

$$x^{2}-6x+9=-4+9$$

Add -4 to 9.

$$x^{2}-6x+9=5$$

Factor x^{2}-6x+9. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-3\right)^{2}=5$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-3\right)^{2}}=\sqrt{5}$$

Simplify.

$$x-3=\sqrt{5} x-3=-\sqrt{5}$$

Add 3 to both sides of the equation.

$$x=\sqrt{5}+3 x=3-\sqrt{5}$$