Solve 1/4x^2+4=x^2

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

Solve for x

$$x = \frac{4 \sqrt{3}}{3} \approx 2.309401077$$

$$x = -\frac{4 \sqrt{3}}{3} \approx -2.309401077$$

Steps by Finding Square Root

Subtract x^{2} from both sides.

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

Combine \frac{1}{4}x^{2} and -x^{2} to get -\frac{3}{4}x^{2}.

$$-\frac{3}{4}x^{2}+4=0$$

Subtract 4 from both sides. Anything subtracted from zero gives its negation.

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

Multiply both sides by -\frac{4}{3}, the reciprocal of -\frac{3}{4}.

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

Multiply -4 and -\frac{4}{3} to get \frac{16}{3}.

$$x^{2}=\frac{16}{3}$$

Take the square root of both sides of the equation.

$$x=\frac{4\sqrt{3}}{3} x=-\frac{4\sqrt{3}}{3}$$

Steps Using the Quadratic Formula

Subtract x^{2} from both sides.

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

Combine \frac{1}{4}x^{2} and -x^{2} to get -\frac{3}{4}x^{2}.

$$-\frac{3}{4}x^{2}+4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -\frac{3}{4} for a, 0 for b, and 4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-\frac{3}{4}\right)\times 4}}{2\left(-\frac{3}{4}\right)}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-\frac{3}{4}\right)\times 4}}{2\left(-\frac{3}{4}\right)}$$

Multiply -4 times -\frac{3}{4}.

$$x=\frac{0±\sqrt{3\times 4}}{2\left(-\frac{3}{4}\right)}$$

Multiply 3 times 4.

$$x=\frac{0±\sqrt{12}}{2\left(-\frac{3}{4}\right)}$$

Take the square root of 12.

$$x=\frac{0±2\sqrt{3}}{2\left(-\frac{3}{4}\right)}$$

Multiply 2 times -\frac{3}{4}.

$$x=\frac{0±2\sqrt{3}}{-\frac{3}{2}}$$

Now solve the equation x=\frac{0±2\sqrt{3}}{-\frac{3}{2}} when ± is plus.

$$x=-\frac{4\sqrt{3}}{3}$$

Now solve the equation x=\frac{0±2\sqrt{3}}{-\frac{3}{2}} when ± is minus.

$$x=\frac{4\sqrt{3}}{3}$$

The equation is now solved.

$$x=-\frac{4\sqrt{3}}{3} x=\frac{4\sqrt{3}}{3}$$