Solve (sigma)(x+sqrt{3})^2-3=0

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

Solve for σ

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

$$x\neq -\sqrt{3}$$

Steps for Solving Linear Equation

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

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

The square of \sqrt{3} is 3.

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

Use the distributive property to multiply \sigma by x^{2}+2x\sqrt{3}+3.

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

Add 3 to both sides. Anything plus zero gives itself.

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

Combine all terms containing \sigma .

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

The equation is in standard form.

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

Divide both sides by x^{2}+2x\sqrt{3}+3.

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

Dividing by x^{2}+2x\sqrt{3}+3 undoes the multiplication by x^{2}+2x\sqrt{3}+3.

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

Divide 3 by x^{2}+2x\sqrt{3}+3.

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

Solve for x (complex solution)

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

$$x=-\sqrt{3}-\sqrt{3}\sigma ^{-\frac{1}{2}}\text{, }\sigma \neq 0$$

Solve for x

$$x=-\sqrt{3}+\sqrt{\frac{3}{\sigma }}$$

$$x=-\sqrt{3}-\sqrt{\frac{3}{\sigma }}\text{, }\sigma >0$$