Solve x^2+x^2=(sqrt{2}+sqrt{6})^2

$$x ^ { 2 } + x ^ { 2 } = ( \sqrt { 2 } + \sqrt { 6 } ) ^ { 2 }$$

Solve for x (complex solution)

$$x=-\left(\sqrt{3}+1\right)\approx -2.732050808$$

$$x=\sqrt{3}+1\approx 2.732050808$$

Steps by Finding Square Root

Combine x^{2} and x^{2} to get 2x^{2}.

$$2x^{2}=\left(\sqrt{2}+\sqrt{6}\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{2}+\sqrt{6}\right)^{2}.

$$2x^{2}=\left(\sqrt{2}\right)^{2}+2\sqrt{2}\sqrt{6}+\left(\sqrt{6}\right)^{2}$$

The square of \sqrt{2} is 2.

$$2x^{2}=2+2\sqrt{2}\sqrt{6}+\left(\sqrt{6}\right)^{2}$$

Factor 6=2\times 3. Rewrite the square root of the product \sqrt{2\times 3} as the product of square roots \sqrt{2}\sqrt{3}.

$$2x^{2}=2+2\sqrt{2}\sqrt{2}\sqrt{3}+\left(\sqrt{6}\right)^{2}$$

Multiply \sqrt{2} and \sqrt{2} to get 2.

$$2x^{2}=2+2\times 2\sqrt{3}+\left(\sqrt{6}\right)^{2}$$

Multiply 2 and 2 to get 4.

$$2x^{2}=2+4\sqrt{3}+\left(\sqrt{6}\right)^{2}$$

The square of \sqrt{6} is 6.

$$2x^{2}=2+4\sqrt{3}+6$$

Add 2 and 6 to get 8.

$$2x^{2}=8+4\sqrt{3}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}=\frac{4\sqrt{3}+8}{2}$$

Divide 8+4\sqrt{3} by 2.

$$x^{2}=2\sqrt{3}+4$$

Take the square root of both sides of the equation.

$$x=\sqrt{3}+1 x=-\left(\sqrt{3}+1\right)$$

Steps Using the Quadratic Formula

Combine x^{2} and x^{2} to get 2x^{2}.

$$2x^{2}=\left(\sqrt{2}+\sqrt{6}\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{2}+\sqrt{6}\right)^{2}.

$$2x^{2}=\left(\sqrt{2}\right)^{2}+2\sqrt{2}\sqrt{6}+\left(\sqrt{6}\right)^{2}$$

The square of \sqrt{2} is 2.

$$2x^{2}=2+2\sqrt{2}\sqrt{6}+\left(\sqrt{6}\right)^{2}$$

Factor 6=2\times 3. Rewrite the square root of the product \sqrt{2\times 3} as the product of square roots \sqrt{2}\sqrt{3}.

$$2x^{2}=2+2\sqrt{2}\sqrt{2}\sqrt{3}+\left(\sqrt{6}\right)^{2}$$

Multiply \sqrt{2} and \sqrt{2} to get 2.

$$2x^{2}=2+2\times 2\sqrt{3}+\left(\sqrt{6}\right)^{2}$$

Multiply 2 and 2 to get 4.

$$2x^{2}=2+4\sqrt{3}+\left(\sqrt{6}\right)^{2}$$

The square of \sqrt{6} is 6.

$$2x^{2}=2+4\sqrt{3}+6$$

Add 2 and 6 to get 8.

$$2x^{2}=8+4\sqrt{3}$$

Subtract 8 from both sides.

$$2x^{2}-8=4\sqrt{3}$$

Subtract 4\sqrt{3} from both sides.

$$2x^{2}-8-4\sqrt{3}=0$$

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.

$$2x^{2}-4\sqrt{3}-8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 0 for b, and -8-4\sqrt{3} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\times 2\left(-4\sqrt{3}-8\right)}}{2\times 2}$$

Square 0.

$$x=\frac{0±\sqrt{-4\times 2\left(-4\sqrt{3}-8\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{0±\sqrt{-8\left(-4\sqrt{3}-8\right)}}{2\times 2}$$

Multiply -8 times -8-4\sqrt{3}.

$$x=\frac{0±\sqrt{32\sqrt{3}+64}}{2\times 2}$$

Take the square root of 64+32\sqrt{3}.

$$x=\frac{0±2^{\frac{5}{2}}\times \frac{\sqrt{2}+\sqrt{6}}{2}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{0±2^{\frac{5}{2}}\times \frac{\sqrt{2}+\sqrt{6}}{2}}{4}$$

Now solve the equation x=\frac{0±2^{\frac{5}{2}}\times \frac{\sqrt{2}+\sqrt{6}}{2}}{4} when ± is plus.

$$x=\sqrt{3}+1$$

Now solve the equation x=\frac{0±2^{\frac{5}{2}}\times \frac{\sqrt{2}+\sqrt{6}}{2}}{4} when ± is minus.

$$x=-\sqrt{3}-1$$

The equation is now solved.

$$x=\sqrt{3}+1 x=-\sqrt{3}-1$$

Solve for x

$$x=\sqrt{3}+1\approx 2.732050808$$

$$x=-\sqrt{3}-1\approx -2.732050808$$