Solve 4x^2=x+4

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

Solve for x

$$x = \frac{\sqrt{65} + 1}{8} \approx 1.132782219$$

$$x=\frac{1-\sqrt{65}}{8}\approx -0.882782219$$

Steps Using the Quadratic Formula

Subtract x from both sides.

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

Subtract 4 from both sides.

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

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

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

Multiply -4 times 4.

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

Multiply -16 times -4.

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

Add 1 to 64.

$$x=\frac{-\left(-1\right)±\sqrt{65}}{2\times 4}$$

The opposite of -1 is 1.

$$x=\frac{1±\sqrt{65}}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{1±\sqrt{65}}{8}$$

Now solve the equation x=\frac{1±\sqrt{65}}{8} when ± is plus. Add 1 to \sqrt{65}.

$$x=\frac{\sqrt{65}+1}{8}$$

Now solve the equation x=\frac{1±\sqrt{65}}{8} when ± is minus. Subtract \sqrt{65} from 1.

$$x=\frac{1-\sqrt{65}}{8}$$

The equation is now solved.

$$x=\frac{\sqrt{65}+1}{8} x=\frac{1-\sqrt{65}}{8}$$

Steps for Completing the Square

Subtract x from both sides.

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

Divide both sides by 4.

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

Dividing by 4 undoes the multiplication by 4.

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

Divide 4 by 4.

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

Divide -\frac{1}{4}, the coefficient of the x term, by 2 to get -\frac{1}{8}. Then add the square of -\frac{1}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{1}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{4}x+\frac{1}{64}=1+\frac{1}{64}$$

Add 1 to \frac{1}{64}.

$$x^{2}-\frac{1}{4}x+\frac{1}{64}=\frac{65}{64}$$

Factor x^{2}-\frac{1}{4}x+\frac{1}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{8}\right)^{2}=\frac{65}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{8}\right)^{2}}=\sqrt{\frac{65}{64}}$$

Simplify.

$$x-\frac{1}{8}=\frac{\sqrt{65}}{8} x-\frac{1}{8}=-\frac{\sqrt{65}}{8}$$

Add \frac{1}{8} to both sides of the equation.

$$x=\frac{\sqrt{65}+1}{8} x=\frac{1-\sqrt{65}}{8}$$