Solve (x+sqrt{5})^2=(xsqrt{5}+1)^2

$$( x + \sqrt { 5 } ) ^ { 2 } = ( x \sqrt { 5 } + 1 ) ^ { 2 }$$

Solve for x

$$x=1$$

$$x=-1$$

Steps by Finding Square Root

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

$$x^{2}+2x\sqrt{5}+\left(\sqrt{5}\right)^{2}=\left(x\sqrt{5}+1\right)^{2}$$

The square of \sqrt{5} is 5.

$$x^{2}+2x\sqrt{5}+5=\left(x\sqrt{5}+1\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x\sqrt{5}+1\right)^{2}.

$$x^{2}+2x\sqrt{5}+5=x^{2}\left(\sqrt{5}\right)^{2}+2x\sqrt{5}+1$$

The square of \sqrt{5} is 5.

$$x^{2}+2x\sqrt{5}+5=x^{2}\times 5+2x\sqrt{5}+1$$

Subtract x^{2}\times 5 from both sides.

$$x^{2}+2x\sqrt{5}+5-x^{2}\times 5=2x\sqrt{5}+1$$

Combine x^{2} and -x^{2}\times 5 to get -4x^{2}.

$$-4x^{2}+2x\sqrt{5}+5=2x\sqrt{5}+1$$

Subtract 2x\sqrt{5} from both sides.

$$-4x^{2}+2x\sqrt{5}+5-2x\sqrt{5}=1$$

Combine 2x\sqrt{5} and -2x\sqrt{5} to get 0.

$$-4x^{2}+5=1$$

Subtract 5 from both sides.

$$-4x^{2}=1-5$$

Subtract 5 from 1 to get -4.

$$-4x^{2}=-4$$

Divide both sides by -4.

$$x^{2}=\frac{-4}{-4}$$

Divide -4 by -4 to get 1.

$$x^{2}=1$$

Take the square root of both sides of the equation.

$$x=1 x=-1$$

Steps Using the Quadratic Formula

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

$$x^{2}+2x\sqrt{5}+\left(\sqrt{5}\right)^{2}=\left(x\sqrt{5}+1\right)^{2}$$

The square of \sqrt{5} is 5.

$$x^{2}+2x\sqrt{5}+5=\left(x\sqrt{5}+1\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x\sqrt{5}+1\right)^{2}.

$$x^{2}+2x\sqrt{5}+5=x^{2}\left(\sqrt{5}\right)^{2}+2x\sqrt{5}+1$$

The square of \sqrt{5} is 5.

$$x^{2}+2x\sqrt{5}+5=x^{2}\times 5+2x\sqrt{5}+1$$

Subtract x^{2}\times 5 from both sides.

$$x^{2}+2x\sqrt{5}+5-x^{2}\times 5=2x\sqrt{5}+1$$

Combine x^{2} and -x^{2}\times 5 to get -4x^{2}.

$$-4x^{2}+2x\sqrt{5}+5=2x\sqrt{5}+1$$

Subtract 2x\sqrt{5} from both sides.

$$-4x^{2}+2x\sqrt{5}+5-2x\sqrt{5}=1$$

Combine 2x\sqrt{5} and -2x\sqrt{5} to get 0.

$$-4x^{2}+5=1$$

Subtract 1 from both sides.

$$-4x^{2}+5-1=0$$

Subtract 1 from 5 to get 4.

$$-4x^{2}+4=0$$

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

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

Square 0.

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

Multiply -4 times -4.

$$x=\frac{0±\sqrt{16\times 4}}{2\left(-4\right)}$$

Multiply 16 times 4.

$$x=\frac{0±\sqrt{64}}{2\left(-4\right)}$$

Take the square root of 64.

$$x=\frac{0±8}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{0±8}{-8}$$

Now solve the equation x=\frac{0±8}{-8} when ± is plus. Divide 8 by -8.

$$x=-1$$

Now solve the equation x=\frac{0±8}{-8} when ± is minus. Divide -8 by -8.

$$x=1$$

The equation is now solved.

$$x=-1 x=1$$