Solve 4x=4x^2-8x+4

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

Solve for x

$$x = \frac{\sqrt{5} + 3}{2} \approx 2.618033989$$

$$x=\frac{3-\sqrt{5}}{2}\approx 0.381966011$$

Steps Using the Quadratic Formula

Subtract 4x^{2} from both sides.

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

Add 8x to both sides.

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

Combine 4x and 8x to get 12x.

$$12x-4x^{2}=4$$

Subtract 4 from both sides.

$$12x-4x^{2}-4=0$$

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.

$$-4x^{2}+12x-4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -4 for a, 12 for b, and -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square 12.

$$x=\frac{-12±\sqrt{144-4\left(-4\right)\left(-4\right)}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$x=\frac{-12±\sqrt{144+16\left(-4\right)}}{2\left(-4\right)}$$

Multiply 16 times -4.

$$x=\frac{-12±\sqrt{144-64}}{2\left(-4\right)}$$

Add 144 to -64.

$$x=\frac{-12±\sqrt{80}}{2\left(-4\right)}$$

Take the square root of 80.

$$x=\frac{-12±4\sqrt{5}}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{-12±4\sqrt{5}}{-8}$$

Now solve the equation x=\frac{-12±4\sqrt{5}}{-8} when ± is plus. Add -12 to 4\sqrt{5}.

$$x=\frac{4\sqrt{5}-12}{-8}$$

Divide -12+4\sqrt{5} by -8.

$$x=\frac{3-\sqrt{5}}{2}$$

Now solve the equation x=\frac{-12±4\sqrt{5}}{-8} when ± is minus. Subtract 4\sqrt{5} from -12.

$$x=\frac{-4\sqrt{5}-12}{-8}$$

Divide -12-4\sqrt{5} by -8.

$$x=\frac{\sqrt{5}+3}{2}$$

The equation is now solved.

$$x=\frac{3-\sqrt{5}}{2} x=\frac{\sqrt{5}+3}{2}$$

Steps for Completing the Square

Subtract 4x^{2} from both sides.

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

Add 8x to both sides.

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

Combine 4x and 8x to get 12x.

$$12x-4x^{2}=4$$

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.

$$-4x^{2}+12x=4$$

Divide both sides by -4.

$$\frac{-4x^{2}+12x}{-4}=\frac{4}{-4}$$

Dividing by -4 undoes the multiplication by -4.

$$x^{2}+\frac{12}{-4}x=\frac{4}{-4}$$

Divide 12 by -4.

$$x^{2}-3x=\frac{4}{-4}$$

Divide 4 by -4.

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

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

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

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-3x+\frac{9}{4}=-1+\frac{9}{4}$$

Add -1 to \frac{9}{4}.

$$x^{2}-3x+\frac{9}{4}=\frac{5}{4}$$

Factor x^{2}-3x+\frac{9}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{3}{2}\right)^{2}=\frac{5}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{2}\right)^{2}}=\sqrt{\frac{5}{4}}$$

Simplify.

$$x-\frac{3}{2}=\frac{\sqrt{5}}{2} x-\frac{3}{2}=-\frac{\sqrt{5}}{2}$$

Add \frac{3}{2} to both sides of the equation.

$$x=\frac{\sqrt{5}+3}{2} x=\frac{3-\sqrt{5}}{2}$$