Solve (x/2)^2+2=(frac{sqrt{3}x}{2})^2

$$( \frac { x } { 2 } ) ^ { 2 } + 2 = ( \frac { \sqrt { 3 } x } { 2 } ) ^ { 2 }$$

Solve for x

$$x=2$$

$$x=-2$$

Steps by Finding Square Root

To raise \frac{x}{2} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{x^{2}}{2^{2}}+2=\left(\frac{\sqrt{3}x}{2}\right)^{2}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 2 times \frac{2^{2}}{2^{2}}.

$$\frac{x^{2}}{2^{2}}+\frac{2\times 2^{2}}{2^{2}}=\left(\frac{\sqrt{3}x}{2}\right)^{2}$$

Since \frac{x^{2}}{2^{2}} and \frac{2\times 2^{2}}{2^{2}} have the same denominator, add them by adding their numerators.

$$\frac{x^{2}+2\times 2^{2}}{2^{2}}=\left(\frac{\sqrt{3}x}{2}\right)^{2}$$

Do the multiplications in x^{2}+2\times 2^{2}.

$$\frac{x^{2}+8}{2^{2}}=\left(\frac{\sqrt{3}x}{2}\right)^{2}$$

To raise \frac{\sqrt{3}x}{2} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{x^{2}+8}{2^{2}}=\frac{\left(\sqrt{3}x\right)^{2}}{2^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{x^{2}+8}{4}=\frac{\left(\sqrt{3}x\right)^{2}}{2^{2}}$$

Divide each term of x^{2}+8 by 4 to get \frac{1}{4}x^{2}+2.

$$\frac{1}{4}x^{2}+2=\frac{\left(\sqrt{3}x\right)^{2}}{2^{2}}$$

Expand \left(\sqrt{3}x\right)^{2}.

$$\frac{1}{4}x^{2}+2=\frac{\left(\sqrt{3}\right)^{2}x^{2}}{2^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{1}{4}x^{2}+2=\frac{3x^{2}}{2^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{1}{4}x^{2}+2=\frac{3x^{2}}{4}$$

Subtract \frac{3x^{2}}{4} from both sides.

$$\frac{1}{4}x^{2}+2-\frac{3x^{2}}{4}=0$$

Subtract 2 from both sides. Anything subtracted from zero gives its negation.

$$\frac{1}{4}x^{2}-\frac{3x^{2}}{4}=-2$$

Multiply both sides of the equation by 4.

$$1x^{2}-3x^{2}=-8$$

Reorder the terms.

$$-3x^{2}+x^{2}=-8$$

Combine -3x^{2} and x^{2} to get -2x^{2}.

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

Divide both sides by -2.

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

Divide -8 by -2 to get 4.

$$x^{2}=4$$

Take the square root of both sides of the equation.

$$x=2 x=-2$$

Steps Using the Quadratic Formula

To raise \frac{x}{2} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{x^{2}}{2^{2}}+2=\left(\frac{\sqrt{3}x}{2}\right)^{2}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 2 times \frac{2^{2}}{2^{2}}.

$$\frac{x^{2}}{2^{2}}+\frac{2\times 2^{2}}{2^{2}}=\left(\frac{\sqrt{3}x}{2}\right)^{2}$$

Since \frac{x^{2}}{2^{2}} and \frac{2\times 2^{2}}{2^{2}} have the same denominator, add them by adding their numerators.

$$\frac{x^{2}+2\times 2^{2}}{2^{2}}=\left(\frac{\sqrt{3}x}{2}\right)^{2}$$

Do the multiplications in x^{2}+2\times 2^{2}.

$$\frac{x^{2}+8}{2^{2}}=\left(\frac{\sqrt{3}x}{2}\right)^{2}$$

To raise \frac{\sqrt{3}x}{2} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{x^{2}+8}{2^{2}}=\frac{\left(\sqrt{3}x\right)^{2}}{2^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{x^{2}+8}{4}=\frac{\left(\sqrt{3}x\right)^{2}}{2^{2}}$$

Divide each term of x^{2}+8 by 4 to get \frac{1}{4}x^{2}+2.

$$\frac{1}{4}x^{2}+2=\frac{\left(\sqrt{3}x\right)^{2}}{2^{2}}$$

Expand \left(\sqrt{3}x\right)^{2}.

$$\frac{1}{4}x^{2}+2=\frac{\left(\sqrt{3}\right)^{2}x^{2}}{2^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{1}{4}x^{2}+2=\frac{3x^{2}}{2^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{1}{4}x^{2}+2=\frac{3x^{2}}{4}$$

Subtract \frac{3x^{2}}{4} from both sides.

$$\frac{1}{4}x^{2}+2-\frac{3x^{2}}{4}=0$$

Multiply both sides of the equation by 4.

$$1x^{2}+8-3x^{2}=0$$

Reorder the terms.

$$-3x^{2}+x^{2}+8=0$$

Combine -3x^{2} and x^{2} to get -2x^{2}.

$$-2x^{2}+8=0$$

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

$$x=\frac{0±\sqrt{0^{2}-4\left(-2\right)\times 8}}{2\left(-2\right)}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-2\right)\times 8}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{0±\sqrt{8\times 8}}{2\left(-2\right)}$$

Multiply 8 times 8.

$$x=\frac{0±\sqrt{64}}{2\left(-2\right)}$$

Take the square root of 64.

$$x=\frac{0±8}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{0±8}{-4}$$

Now solve the equation x=\frac{0±8}{-4} when ± is plus. Divide 8 by -4.

$$x=-2$$

Now solve the equation x=\frac{0±8}{-4} when ± is minus. Divide -8 by -4.

$$x=2$$

The equation is now solved.

$$x=-2 x=2$$