Solve {left(x/2right)}^2+3^2=left(2xright)^2

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

Solve for x

$$x = \frac{2 \sqrt{15}}{5} \approx 1.549193338$$

$$x = -\frac{2 \sqrt{15}}{5} \approx -1.549193338$$

Steps by Finding Square Root

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

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

Calculate 3 to the power of 2 and get 9.

$$\frac{x^{2}}{2^{2}}+9=\left(2x\right)^{2}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 9 times \frac{2^{2}}{2^{2}}.

$$\frac{x^{2}}{2^{2}}+\frac{9\times 2^{2}}{2^{2}}=\left(2x\right)^{2}$$

Since \frac{x^{2}}{2^{2}} and \frac{9\times 2^{2}}{2^{2}} have the same denominator, add them by adding their numerators.

$$\frac{x^{2}+9\times 2^{2}}{2^{2}}=\left(2x\right)^{2}$$

Do the multiplications in x^{2}+9\times 2^{2}.

$$\frac{x^{2}+36}{2^{2}}=\left(2x\right)^{2}$$

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

$$\frac{x^{2}+36}{2^{2}}=2^{2}x^{2}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{x^{2}+36}{2^{2}}=4x^{2}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{x^{2}+36}{4}=4x^{2}$$

Divide each term of x^{2}+36 by 4 to get \frac{1}{4}x^{2}+9.

$$\frac{1}{4}x^{2}+9=4x^{2}$$

Subtract 4x^{2} from both sides.

$$\frac{1}{4}x^{2}+9-4x^{2}=0$$

Combine \frac{1}{4}x^{2} and -4x^{2} to get -\frac{15}{4}x^{2}.

$$-\frac{15}{4}x^{2}+9=0$$

Subtract 9 from both sides. Anything subtracted from zero gives its negation.

$$-\frac{15}{4}x^{2}=-9$$

Multiply both sides by -\frac{4}{15}, the reciprocal of -\frac{15}{4}.

$$x^{2}=-9\left(-\frac{4}{15}\right)$$

Multiply -9 and -\frac{4}{15} to get \frac{12}{5}.

$$x^{2}=\frac{12}{5}$$

Take the square root of both sides of the equation.

$$x=\frac{2\sqrt{15}}{5} x=-\frac{2\sqrt{15}}{5}$$

Steps Using the Quadratic Formula

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

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

Calculate 3 to the power of 2 and get 9.

$$\frac{x^{2}}{2^{2}}+9=\left(2x\right)^{2}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 9 times \frac{2^{2}}{2^{2}}.

$$\frac{x^{2}}{2^{2}}+\frac{9\times 2^{2}}{2^{2}}=\left(2x\right)^{2}$$

Since \frac{x^{2}}{2^{2}} and \frac{9\times 2^{2}}{2^{2}} have the same denominator, add them by adding their numerators.

$$\frac{x^{2}+9\times 2^{2}}{2^{2}}=\left(2x\right)^{2}$$

Do the multiplications in x^{2}+9\times 2^{2}.

$$\frac{x^{2}+36}{2^{2}}=\left(2x\right)^{2}$$

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

$$\frac{x^{2}+36}{2^{2}}=2^{2}x^{2}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{x^{2}+36}{2^{2}}=4x^{2}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{x^{2}+36}{4}=4x^{2}$$

Divide each term of x^{2}+36 by 4 to get \frac{1}{4}x^{2}+9.

$$\frac{1}{4}x^{2}+9=4x^{2}$$

Subtract 4x^{2} from both sides.

$$\frac{1}{4}x^{2}+9-4x^{2}=0$$

Combine \frac{1}{4}x^{2} and -4x^{2} to get -\frac{15}{4}x^{2}.

$$-\frac{15}{4}x^{2}+9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -\frac{15}{4} for a, 0 for b, and 9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-\frac{15}{4}\right)\times 9}}{2\left(-\frac{15}{4}\right)}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-\frac{15}{4}\right)\times 9}}{2\left(-\frac{15}{4}\right)}$$

Multiply -4 times -\frac{15}{4}.

$$x=\frac{0±\sqrt{15\times 9}}{2\left(-\frac{15}{4}\right)}$$

Multiply 15 times 9.

$$x=\frac{0±\sqrt{135}}{2\left(-\frac{15}{4}\right)}$$

Take the square root of 135.

$$x=\frac{0±3\sqrt{15}}{2\left(-\frac{15}{4}\right)}$$

Multiply 2 times -\frac{15}{4}.

$$x=\frac{0±3\sqrt{15}}{-\frac{15}{2}}$$

Now solve the equation x=\frac{0±3\sqrt{15}}{-\frac{15}{2}} when ± is plus.

$$x=-\frac{2\sqrt{15}}{5}$$

Now solve the equation x=\frac{0±3\sqrt{15}}{-\frac{15}{2}} when ± is minus.

$$x=\frac{2\sqrt{15}}{5}$$

The equation is now solved.

$$x=-\frac{2\sqrt{15}}{5} x=\frac{2\sqrt{15}}{5}$$