Solve (sqrt{3}-1)^2=x^2+1

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

Solve for x (complex solution)

$$x=-\frac{\left(\sqrt{6}-\sqrt{2}\right)\sqrt[4]{3}i}{2}\approx -0-0.681250039i$$

$$x=\frac{\left(\sqrt{6}-\sqrt{2}\right)\sqrt[4]{3}i}{2}\approx 0.681250039i$$

Steps by Finding Square Root

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{3}-1\right)^{2}.

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

The square of \sqrt{3} is 3.

$$3-2\sqrt{3}+1=x^{2}+1$$

Add 3 and 1 to get 4.

$$4-2\sqrt{3}=x^{2}+1$$

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

$$x^{2}+1=4-2\sqrt{3}$$

Subtract 1 from both sides.

$$x^{2}=4-2\sqrt{3}-1$$

Subtract 1 from 4 to get 3.

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

The equation is now solved.

$$x=\frac{\sqrt[4]{3}\left(\sqrt{6}i-\sqrt{2}i\right)}{2} x=-\frac{\sqrt[4]{3}\left(\sqrt{6}i-\sqrt{2}i\right)}{2}$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{3}-1\right)^{2}.

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

The square of \sqrt{3} is 3.

$$3-2\sqrt{3}+1=x^{2}+1$$

Add 3 and 1 to get 4.

$$4-2\sqrt{3}=x^{2}+1$$

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

$$x^{2}+1=4-2\sqrt{3}$$

Subtract 4 from both sides.

$$x^{2}+1-4=-2\sqrt{3}$$

Subtract 4 from 1 to get -3.

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

Add 2\sqrt{3} to both sides.

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

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.

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

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

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

Square 0.

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

Multiply -4 times -3+2\sqrt{3}.

$$x=\frac{0±\sqrt{12-8\sqrt{3}}}{2}$$

Take the square root of 12-8\sqrt{3}.

$$x=\frac{0±2\sqrt[4]{3}\times \frac{\sqrt{6}i-\sqrt{2}i}{2}}{2}$$

Now solve the equation x=\frac{0±2\sqrt[4]{3}\times \frac{\sqrt{6}i-\sqrt{2}i}{2}}{2} when ± is plus.

$$x=\frac{\sqrt[4]{3}\left(\sqrt{6}i-\sqrt{2}i\right)}{2}$$

Now solve the equation x=\frac{0±2\sqrt[4]{3}\times \frac{\sqrt{6}i-\sqrt{2}i}{2}}{2} when ± is minus.

$$x=-\frac{\sqrt[4]{3}\left(\sqrt{6}i-\sqrt{2}i\right)}{2}$$

The equation is now solved.

$$x=\frac{\sqrt[4]{3}\left(\sqrt{6}i-\sqrt{2}i\right)}{2} x=-\frac{\sqrt[4]{3}\left(\sqrt{6}i-\sqrt{2}i\right)}{2}$$