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

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

Solve for x

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

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

Steps by Finding Square Root

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2+\sqrt{5}\right)^{2}.

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

The square of \sqrt{5} is 5.

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

Add 4 and 5 to get 9.

$$x^{2}=9+4\sqrt{5}+\left(2-\sqrt{5}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2-\sqrt{5}\right)^{2}.

$$x^{2}=9+4\sqrt{5}+4-4\sqrt{5}+\left(\sqrt{5}\right)^{2}$$

The square of \sqrt{5} is 5.

$$x^{2}=9+4\sqrt{5}+4-4\sqrt{5}+5$$

Add 4 and 5 to get 9.

$$x^{2}=9+4\sqrt{5}+9-4\sqrt{5}$$

Add 9 and 9 to get 18.

$$x^{2}=18+4\sqrt{5}-4\sqrt{5}$$

Combine 4\sqrt{5} and -4\sqrt{5} to get 0.

$$x^{2}=18$$

Take the square root of both sides of the equation.

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

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2+\sqrt{5}\right)^{2}.

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

The square of \sqrt{5} is 5.

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

Add 4 and 5 to get 9.

$$x^{2}=9+4\sqrt{5}+\left(2-\sqrt{5}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2-\sqrt{5}\right)^{2}.

$$x^{2}=9+4\sqrt{5}+4-4\sqrt{5}+\left(\sqrt{5}\right)^{2}$$

The square of \sqrt{5} is 5.

$$x^{2}=9+4\sqrt{5}+4-4\sqrt{5}+5$$

Add 4 and 5 to get 9.

$$x^{2}=9+4\sqrt{5}+9-4\sqrt{5}$$

Add 9 and 9 to get 18.

$$x^{2}=18+4\sqrt{5}-4\sqrt{5}$$

Combine 4\sqrt{5} and -4\sqrt{5} to get 0.

$$x^{2}=18$$

Subtract 18 from both sides.

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

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.

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