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

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

Solve for x (complex solution)

$$x=\frac{\sqrt{2}+\sqrt{6}+2i\sqrt{\sqrt{3}+1}}{2}\approx 1.931851653+1.65289165i$$

$$x=\frac{-2i\sqrt{\sqrt{3}+1}+\sqrt{2}+\sqrt{6}}{2}\approx 1.931851653-1.65289165i$$

Steps Using the Quadratic Formula

Square \sqrt{6}+\sqrt{2}-x.

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

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}.

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

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

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

Multiply 2 and 2 to get 4.

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

The square of \sqrt{2} is 2.

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

The square of \sqrt{6} is 6.

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

Add 2 and 6 to get 8.

$$x^{2}+x^{2}-2\sqrt{2}x-2\sqrt{6}x+4\sqrt{3}+8=2$$

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

$$2x^{2}-2\sqrt{2}x-2\sqrt{6}x+4\sqrt{3}+8=2$$

Subtract 2 from both sides.

$$2x^{2}-2\sqrt{2}x-2\sqrt{6}x+4\sqrt{3}+8-2=0$$

Subtract 2 from 8 to get 6.

$$2x^{2}-2\sqrt{2}x-2\sqrt{6}x+4\sqrt{3}+6=0$$

Combine all terms containing x.

$$2x^{2}+\left(-2\sqrt{2}-2\sqrt{6}\right)x+4\sqrt{3}+6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -2\sqrt{2}-2\sqrt{6} for b, and 4\sqrt{3}+6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square -2\sqrt{2}-2\sqrt{6}.

$$x=\frac{-\left(-2\sqrt{2}-2\sqrt{6}\right)±\sqrt{16\sqrt{3}+32-4\times 2\left(4\sqrt{3}+6\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-2\sqrt{2}-2\sqrt{6}\right)±\sqrt{16\sqrt{3}+32-8\left(4\sqrt{3}+6\right)}}{2\times 2}$$

Multiply -8 times 4\sqrt{3}+6.

$$x=\frac{-\left(-2\sqrt{2}-2\sqrt{6}\right)±\sqrt{16\sqrt{3}+32-32\sqrt{3}-48}}{2\times 2}$$

Add 32+16\sqrt{3} to -32\sqrt{3}-48.

$$x=\frac{-\left(-2\sqrt{2}-2\sqrt{6}\right)±\sqrt{-16\sqrt{3}-16}}{2\times 2}$$

Take the square root of -16-16\sqrt{3}.

$$x=\frac{-\left(-2\sqrt{2}-2\sqrt{6}\right)±4i\sqrt{\sqrt{3}+1}}{2\times 2}$$

The opposite of -2\sqrt{2}-2\sqrt{6} is 2\sqrt{2}+2\sqrt{6}.

$$x=\frac{2\sqrt{2}+2\sqrt{6}±4i\sqrt{\sqrt{3}+1}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{2\sqrt{2}+2\sqrt{6}±4i\sqrt{\sqrt{3}+1}}{4}$$

Now solve the equation x=\frac{2\sqrt{2}+2\sqrt{6}±4i\sqrt{\sqrt{3}+1}}{4} when ± is plus. Add 2\sqrt{2}+2\sqrt{6} to 4i\sqrt{1+\sqrt{3}}.

$$x=\frac{2\sqrt{2}+2\sqrt{6}+4i\sqrt{\sqrt{3}+1}}{4}$$

Divide 2\sqrt{2}+2\sqrt{6}+4i\sqrt{1+\sqrt{3}} by 4.

$$x=\frac{\sqrt{2}}{2}+\frac{\sqrt{6}}{2}+i\sqrt{\sqrt{3}+1}$$

Now solve the equation x=\frac{2\sqrt{2}+2\sqrt{6}±4i\sqrt{\sqrt{3}+1}}{4} when ± is minus. Subtract 4i\sqrt{1+\sqrt{3}} from 2\sqrt{2}+2\sqrt{6}.

$$x=\frac{-4i\sqrt{\sqrt{3}+1}+2\sqrt{2}+2\sqrt{6}}{4}$$

Divide 2\sqrt{2}+2\sqrt{6}-4i\sqrt{1+\sqrt{3}} by 4.

$$x=-i\sqrt{\sqrt{3}+1}+\frac{\sqrt{2}}{2}+\frac{\sqrt{6}}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{2}}{2}+\frac{\sqrt{6}}{2}+i\sqrt{\sqrt{3}+1} x=-i\sqrt{\sqrt{3}+1}+\frac{\sqrt{2}}{2}+\frac{\sqrt{6}}{2}$$

Steps for Completing the Square

Square \sqrt{6}+\sqrt{2}-x.

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

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}.

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

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

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

Multiply 2 and 2 to get 4.

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

The square of \sqrt{2} is 2.

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

The square of \sqrt{6} is 6.

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

Add 2 and 6 to get 8.

$$x^{2}+x^{2}-2\sqrt{2}x-2\sqrt{6}x+4\sqrt{3}+8=2$$

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

$$2x^{2}-2\sqrt{2}x-2\sqrt{6}x+4\sqrt{3}+8=2$$

Subtract 4\sqrt{3} from both sides.

$$2x^{2}-2\sqrt{2}x-2\sqrt{6}x+8=2-4\sqrt{3}$$

Subtract 8 from both sides.

$$2x^{2}-2\sqrt{2}x-2\sqrt{6}x=2-4\sqrt{3}-8$$

Subtract 8 from 2 to get -6.

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

Combine all terms containing x.

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

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

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

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

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

Divide -2\sqrt{2}-2\sqrt{6} by 2.

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

Divide -6-4\sqrt{3} by 2.

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

Divide -\sqrt{2}-\sqrt{6}, the coefficient of the x term, by 2 to get \frac{-\sqrt{2}-\sqrt{6}}{2}. Then add the square of \frac{-\sqrt{2}-\sqrt{6}}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square \frac{-\sqrt{2}-\sqrt{6}}{2}.

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

Add -3-2\sqrt{3} to 2+\sqrt{3}.

$$x^{2}+\left(-\sqrt{2}-\sqrt{6}\right)x+\sqrt{3}+2=-\sqrt{3}-1$$

Factor x^{2}+\left(-\sqrt{2}-\sqrt{6}\right)x+\sqrt{3}+2. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{-\sqrt{2}-\sqrt{6}}{2}\right)^{2}=-\sqrt{3}-1$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{-\sqrt{2}-\sqrt{6}}{2}\right)^{2}}=\sqrt{-\sqrt{3}-1}$$

Simplify.

$$x+\frac{-\sqrt{2}-\sqrt{6}}{2}=i\sqrt{\sqrt{3}+1} x+\frac{-\sqrt{2}-\sqrt{6}}{2}=-i\sqrt{\sqrt{3}+1}$$

Subtract \frac{-\sqrt{2}-\sqrt{6}}{2} from both sides of the equation.

$$x=\frac{\sqrt{2}}{2}+\frac{\sqrt{6}}{2}+i\sqrt{\sqrt{3}+1} x=-i\sqrt{\sqrt{3}+1}+\frac{\sqrt{2}}{2}+\frac{\sqrt{6}}{2}$$