Solve left(x+1right)^2+left(x-1right)^2=8

$${ \left(x+1 \right) }^{ 2 } + { \left(x-1 \right) }^{ 2 } =8$$

Solve for x

$$x=\sqrt{3}\approx 1.732050808$$

$$x=-\sqrt{3}\approx -1.732050808$$

Steps by Finding Square Root

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

$$x^{2}+2x+1+\left(x-1\right)^{2}=8$$

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

$$x^{2}+2x+1+x^{2}-2x+1=8$$

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

$$2x^{2}+2x+1-2x+1=8$$

Combine 2x and -2x to get 0.

$$2x^{2}+1+1=8$$

Add 1 and 1 to get 2.

$$2x^{2}+2=8$$

Subtract 2 from both sides.

$$2x^{2}=8-2$$

Subtract 2 from 8 to get 6.

$$2x^{2}=6$$

Divide both sides by 2.

$$x^{2}=\frac{6}{2}$$

Divide 6 by 2 to get 3.

$$x^{2}=3$$

Take the square root of both sides of the equation.

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

Steps Using the Quadratic Formula

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

$$x^{2}+2x+1+\left(x-1\right)^{2}=8$$

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

$$x^{2}+2x+1+x^{2}-2x+1=8$$

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

$$2x^{2}+2x+1-2x+1=8$$

Combine 2x and -2x to get 0.

$$2x^{2}+1+1=8$$

Add 1 and 1 to get 2.

$$2x^{2}+2=8$$

Subtract 8 from both sides.

$$2x^{2}+2-8=0$$

Subtract 8 from 2 to get -6.

$$2x^{2}-6=0$$

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

$$x=\frac{0±\sqrt{0^{2}-4\times 2\left(-6\right)}}{2\times 2}$$

Square 0.

$$x=\frac{0±\sqrt{-4\times 2\left(-6\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{0±\sqrt{-8\left(-6\right)}}{2\times 2}$$

Multiply -8 times -6.

$$x=\frac{0±\sqrt{48}}{2\times 2}$$

Take the square root of 48.

$$x=\frac{0±4\sqrt{3}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{0±4\sqrt{3}}{4}$$

Now solve the equation x=\frac{0±4\sqrt{3}}{4} when ± is plus.

$$x=\sqrt{3}$$

Now solve the equation x=\frac{0±4\sqrt{3}}{4} when ± is minus.

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

The equation is now solved.

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