Solve 4x(x-2)=1

$$4 x ( x - 2 ) = 1$$

Solve for x

$$x=\frac{\sqrt{5}}{2}+1\approx 2.118033989$$

$$x=-\frac{\sqrt{5}}{2}+1\approx -0.118033989$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 4x by x-2.

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

Subtract 1 from both sides.

$$4x^{2}-8x-1=0$$

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

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

Square -8.

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

Multiply -4 times 4.

$$x=\frac{-\left(-8\right)±\sqrt{64-16\left(-1\right)}}{2\times 4}$$

Multiply -16 times -1.

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

Add 64 to 16.

$$x=\frac{-\left(-8\right)±\sqrt{80}}{2\times 4}$$

Take the square root of 80.

$$x=\frac{-\left(-8\right)±4\sqrt{5}}{2\times 4}$$

The opposite of -8 is 8.

$$x=\frac{8±4\sqrt{5}}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{8±4\sqrt{5}}{8}$$

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

$$x=\frac{4\sqrt{5}+8}{8}$$

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

$$x=\frac{\sqrt{5}}{2}+1$$

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

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

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

$$x=-\frac{\sqrt{5}}{2}+1$$

The equation is now solved.

$$x=\frac{\sqrt{5}}{2}+1 x=-\frac{\sqrt{5}}{2}+1$$

Steps for Completing the Square

Use the distributive property to multiply 4x by x-2.

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

Divide both sides by 4.

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

Dividing by 4 undoes the multiplication by 4.

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

Divide -8 by 4.

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

Divide -2, the coefficient of the x term, by 2 to get -1. Then add the square of -1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

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

$$x^{2}-2x+1=\frac{5}{4}$$

Factor x^{2}-2x+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-1\right)^{2}=\frac{5}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-1\right)^{2}}=\sqrt{\frac{5}{4}}$$

Simplify.

$$x-1=\frac{\sqrt{5}}{2} x-1=-\frac{\sqrt{5}}{2}$$

Add 1 to both sides of the equation.

$$x=\frac{\sqrt{5}}{2}+1 x=-\frac{\sqrt{5}}{2}+1$$