Solve (x)=1/4x^2

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

Solve for x

$$x=4$$

$$x=0$$

Steps Using Factoring

Subtract \frac{1}{4}x^{2} from both sides.

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

Factor out x.

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

To find equation solutions, solve x=0 and 1-\frac{x}{4}=0.

$$x=0 x=4$$

Steps Using the Quadratic Formula

Subtract \frac{1}{4}x^{2} from both sides.

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

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.

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

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

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

Take the square root of 1^{2}.

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

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

$$x=\frac{-1±1}{-\frac{1}{2}}$$

Now solve the equation x=\frac{-1±1}{-\frac{1}{2}} when ± is plus. Add -1 to 1.

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

Divide 0 by -\frac{1}{2} by multiplying 0 by the reciprocal of -\frac{1}{2}.

$$x=0$$

Now solve the equation x=\frac{-1±1}{-\frac{1}{2}} when ± is minus. Subtract 1 from -1.

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

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

$$x=4$$

The equation is now solved.

$$x=0 x=4$$

Steps for Completing the Square

Subtract \frac{1}{4}x^{2} from both sides.

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

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.

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

Multiply both sides by -4.

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

Dividing by -\frac{1}{4} undoes the multiplication by -\frac{1}{4}.

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

Divide 1 by -\frac{1}{4} by multiplying 1 by the reciprocal of -\frac{1}{4}.

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

Divide 0 by -\frac{1}{4} by multiplying 0 by the reciprocal of -\frac{1}{4}.

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

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

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

Square -2.

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

Factor x^{2}-4x+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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-2=2 x-2=-2$$

Add 2 to both sides of the equation.

$$x=4 x=0$$