Solve x^2=4x+1/4

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

Solve for x

$$x = \frac{\sqrt{2} + 1}{2} \approx 1.207106781$$

$$x=\frac{1-\sqrt{2}}{2}\approx -0.207106781$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 4.

$$4x^{2}=4x+1$$

Subtract 4x from both sides.

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

Subtract 1 from both sides.

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

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

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

Square -4.

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

Multiply -4 times 4.

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

Multiply -16 times -1.

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

Add 16 to 16.

$$x=\frac{-\left(-4\right)±\sqrt{32}}{2\times 4}$$

Take the square root of 32.

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

The opposite of -4 is 4.

$$x=\frac{4±4\sqrt{2}}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{4±4\sqrt{2}}{8}$$

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

$$x=\frac{4\sqrt{2}+4}{8}$$

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

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

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

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

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

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

The equation is now solved.

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

Steps for Completing the Square

Multiply both sides of the equation by 4.

$$4x^{2}=4x+1$$

Subtract 4x from both sides.

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

Divide both sides by 4.

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

Dividing by 4 undoes the multiplication by 4.

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

Divide -4 by 4.

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

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

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

Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.

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

Add \frac{1}{4} to \frac{1}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

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

Factor x^{2}-x+\frac{1}{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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Add \frac{1}{2} to both sides of the equation.

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