Solve x^2=4^2+(2x/5)^2

$$x ^ { 2 } = 4 ^ { 2 } + ( \frac { 2 x } { 5 } ) ^ { 2 }$$

Solve for x

$$x = \frac{20 \sqrt{21}}{21} \approx 4.364357805$$

$$x = -\frac{20 \sqrt{21}}{21} \approx -4.364357805$$

Steps by Finding Square Root

Calculate 4 to the power of 2 and get 16.

$$x^{2}=16+\left(\frac{2x}{5}\right)^{2}$$

To raise \frac{2x}{5} to a power, raise both numerator and denominator to the power and then divide.

$$x^{2}=16+\frac{\left(2x\right)^{2}}{5^{2}}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 16 times \frac{5^{2}}{5^{2}}.

$$x^{2}=\frac{16\times 5^{2}}{5^{2}}+\frac{\left(2x\right)^{2}}{5^{2}}$$

Since \frac{16\times 5^{2}}{5^{2}} and \frac{\left(2x\right)^{2}}{5^{2}} have the same denominator, add them by adding their numerators.

$$x^{2}=\frac{16\times 5^{2}+\left(2x\right)^{2}}{5^{2}}$$

Do the multiplications in 16\times 5^{2}+\left(2x\right)^{2}.

$$x^{2}=\frac{400+\left(2x\right)^{2}}{5^{2}}$$

Combine like terms in 400+\left(2x\right)^{2}.

$$x^{2}=\frac{400+4x^{2}}{5^{2}}$$

Calculate 5 to the power of 2 and get 25.

$$x^{2}=\frac{400+4x^{2}}{25}$$

Divide each term of 400+4x^{2} by 25 to get 16+\frac{4}{25}x^{2}.

$$x^{2}=16+\frac{4}{25}x^{2}$$

Subtract \frac{4}{25}x^{2} from both sides.

$$x^{2}-\frac{4}{25}x^{2}=16$$

Combine x^{2} and -\frac{4}{25}x^{2} to get \frac{21}{25}x^{2}.

$$\frac{21}{25}x^{2}=16$$

Multiply both sides by \frac{25}{21}, the reciprocal of \frac{21}{25}.

$$x^{2}=16\times \frac{25}{21}$$

Multiply 16 and \frac{25}{21} to get \frac{400}{21}.

$$x^{2}=\frac{400}{21}$$

Take the square root of both sides of the equation.

$$x=\frac{20\sqrt{21}}{21} x=-\frac{20\sqrt{21}}{21}$$

Steps Using the Quadratic Formula

Calculate 4 to the power of 2 and get 16.

$$x^{2}=16+\left(\frac{2x}{5}\right)^{2}$$

To raise \frac{2x}{5} to a power, raise both numerator and denominator to the power and then divide.

$$x^{2}=16+\frac{\left(2x\right)^{2}}{5^{2}}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 16 times \frac{5^{2}}{5^{2}}.

$$x^{2}=\frac{16\times 5^{2}}{5^{2}}+\frac{\left(2x\right)^{2}}{5^{2}}$$

Since \frac{16\times 5^{2}}{5^{2}} and \frac{\left(2x\right)^{2}}{5^{2}} have the same denominator, add them by adding their numerators.

$$x^{2}=\frac{16\times 5^{2}+\left(2x\right)^{2}}{5^{2}}$$

Do the multiplications in 16\times 5^{2}+\left(2x\right)^{2}.

$$x^{2}=\frac{400+\left(2x\right)^{2}}{5^{2}}$$

Combine like terms in 400+\left(2x\right)^{2}.

$$x^{2}=\frac{400+4x^{2}}{5^{2}}$$

Calculate 5 to the power of 2 and get 25.

$$x^{2}=\frac{400+4x^{2}}{25}$$

Divide each term of 400+4x^{2} by 25 to get 16+\frac{4}{25}x^{2}.

$$x^{2}=16+\frac{4}{25}x^{2}$$

Subtract 16 from both sides.

$$x^{2}-16=\frac{4}{25}x^{2}$$

Subtract \frac{4}{25}x^{2} from both sides.

$$x^{2}-16-\frac{4}{25}x^{2}=0$$

Combine x^{2} and -\frac{4}{25}x^{2} to get \frac{21}{25}x^{2}.

$$\frac{21}{25}x^{2}-16=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{21}{25} for a, 0 for b, and -16 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\times \frac{21}{25}\left(-16\right)}}{2\times \frac{21}{25}}$$

Square 0.

$$x=\frac{0±\sqrt{-4\times \frac{21}{25}\left(-16\right)}}{2\times \frac{21}{25}}$$

Multiply -4 times \frac{21}{25}.

$$x=\frac{0±\sqrt{-\frac{84}{25}\left(-16\right)}}{2\times \frac{21}{25}}$$

Multiply -\frac{84}{25} times -16.

$$x=\frac{0±\sqrt{\frac{1344}{25}}}{2\times \frac{21}{25}}$$

Take the square root of \frac{1344}{25}.

$$x=\frac{0±\frac{8\sqrt{21}}{5}}{2\times \frac{21}{25}}$$

Multiply 2 times \frac{21}{25}.

$$x=\frac{0±\frac{8\sqrt{21}}{5}}{\frac{42}{25}}$$

Now solve the equation x=\frac{0±\frac{8\sqrt{21}}{5}}{\frac{42}{25}} when ± is plus.

$$x=\frac{20\sqrt{21}}{21}$$

Now solve the equation x=\frac{0±\frac{8\sqrt{21}}{5}}{\frac{42}{25}} when ± is minus.

$$x=-\frac{20\sqrt{21}}{21}$$

The equation is now solved.

$$x=\frac{20\sqrt{21}}{21} x=-\frac{20\sqrt{21}}{21}$$