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

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

Solve for x

$$x=3\sqrt{2}\approx 4.242640687$$

$$x=-3\sqrt{2}\approx -4.242640687$$

Steps by Finding Square Root

Expand \left(2\sqrt{5}\right)^{2}.

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

Calculate 2 to the power of 2 and get 4.

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

The square of \sqrt{5} is 5.

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

Multiply 4 and 5 to get 20.

$$20=\left(\sqrt{2}\right)^{2}+x^{2}$$

The square of \sqrt{2} is 2.

$$20=2+x^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$2+x^{2}=20$$

Subtract 2 from both sides.

$$x^{2}=20-2$$

Subtract 2 from 20 to get 18.

$$x^{2}=18$$

Take the square root of both sides of the equation.

$$x=3\sqrt{2} x=-3\sqrt{2}$$

Steps Using the Quadratic Formula

Expand \left(2\sqrt{5}\right)^{2}.

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

Calculate 2 to the power of 2 and get 4.

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

The square of \sqrt{5} is 5.

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

Multiply 4 and 5 to get 20.

$$20=\left(\sqrt{2}\right)^{2}+x^{2}$$

The square of \sqrt{2} is 2.

$$20=2+x^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$2+x^{2}=20$$

Subtract 20 from both sides.

$$2+x^{2}-20=0$$

Subtract 20 from 2 to get -18.

$$-18+x^{2}=0$$

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.

$$x^{2}-18=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -18 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-18\right)}}{2}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-18\right)}}{2}$$

Multiply -4 times -18.

$$x=\frac{0±\sqrt{72}}{2}$$

Take the square root of 72.

$$x=\frac{0±6\sqrt{2}}{2}$$

Now solve the equation x=\frac{0±6\sqrt{2}}{2} when ± is plus.

$$x=3\sqrt{2}$$

Now solve the equation x=\frac{0±6\sqrt{2}}{2} when ± is minus.

$$x=-3\sqrt{2}$$

The equation is now solved.

$$x=3\sqrt{2} x=-3\sqrt{2}$$