Solve {left(sqrt{5}right)}^2+x^2={left(2sqrt{2}right)}^2

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

Solve for x

$$x=\sqrt{3}\approx 1.732050808$$

$$x=-\sqrt{3}\approx -1.732050808$$

Steps by Finding Square Root

The square of \sqrt{5} is 5.

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

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

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

Calculate 2 to the power of 2 and get 4.

$$5+x^{2}=4\left(\sqrt{2}\right)^{2}$$

The square of \sqrt{2} is 2.

$$5+x^{2}=4\times 2$$

Multiply 4 and 2 to get 8.

$$5+x^{2}=8$$

Subtract 5 from both sides.

$$x^{2}=8-5$$

Subtract 5 from 8 to get 3.

$$x^{2}=3$$

Take the square root of both sides of the equation.

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

Steps Using the Quadratic Formula

The square of \sqrt{5} is 5.

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

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

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

Calculate 2 to the power of 2 and get 4.

$$5+x^{2}=4\left(\sqrt{2}\right)^{2}$$

The square of \sqrt{2} is 2.

$$5+x^{2}=4\times 2$$

Multiply 4 and 2 to get 8.

$$5+x^{2}=8$$

Subtract 8 from both sides.

$$5+x^{2}-8=0$$

Subtract 8 from 5 to get -3.

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

Quadratic equations like this one, with an x^{2} term but no x term, can still be solved using the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}, once they are put in standard form: ax^{2}+bx+c=0.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square 0.

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

Multiply -4 times -3.

$$x=\frac{0±\sqrt{12}}{2}$$

Take the square root of 12.

$$x=\frac{0±2\sqrt{3}}{2}$$

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

$$x=\sqrt{3}$$

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

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

The equation is now solved.

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