Solve (x+1)^2+x^2+(x-1)^2=590

$$( x + 1 ) ^ { 2 } + x ^ { 2 } + ( x - 1 ) ^ { 2 } = 590$$

Solve for x

$$x=-14$$

$$x=14$$

Steps Using Difference of Squares

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

$$x^{2}+2x+1+x^{2}+\left(x-1\right)^{2}=590$$

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

$$2x^{2}+2x+1+\left(x-1\right)^{2}=590$$

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

$$2x^{2}+2x+1+x^{2}-2x+1=590$$

Combine 2x^{2} and x^{2} to get 3x^{2}.

$$3x^{2}+2x+1-2x+1=590$$

Combine 2x and -2x to get 0.

$$3x^{2}+1+1=590$$

Add 1 and 1 to get 2.

$$3x^{2}+2=590$$

Subtract 590 from both sides.

$$3x^{2}+2-590=0$$

Subtract 590 from 2 to get -588.

$$3x^{2}-588=0$$

Divide both sides by 3.

$$x^{2}-196=0$$

Consider x^{2}-196. Rewrite x^{2}-196 as x^{2}-14^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(x-14\right)\left(x+14\right)=0$$

To find equation solutions, solve x-14=0 and x+14=0.

$$x=14 x=-14$$

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}.

$$x^{2}+2x+1+x^{2}+\left(x-1\right)^{2}=590$$

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

$$2x^{2}+2x+1+\left(x-1\right)^{2}=590$$

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

$$2x^{2}+2x+1+x^{2}-2x+1=590$$

Combine 2x^{2} and x^{2} to get 3x^{2}.

$$3x^{2}+2x+1-2x+1=590$$

Combine 2x and -2x to get 0.

$$3x^{2}+1+1=590$$

Add 1 and 1 to get 2.

$$3x^{2}+2=590$$

Subtract 2 from both sides.

$$3x^{2}=590-2$$

Subtract 2 from 590 to get 588.

$$3x^{2}=588$$

Divide both sides by 3.

$$x^{2}=\frac{588}{3}$$

Divide 588 by 3 to get 196.

$$x^{2}=196$$

Take the square root of both sides of the equation.

$$x=14 x=-14$$

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}.

$$x^{2}+2x+1+x^{2}+\left(x-1\right)^{2}=590$$

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

$$2x^{2}+2x+1+\left(x-1\right)^{2}=590$$

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

$$2x^{2}+2x+1+x^{2}-2x+1=590$$

Combine 2x^{2} and x^{2} to get 3x^{2}.

$$3x^{2}+2x+1-2x+1=590$$

Combine 2x and -2x to get 0.

$$3x^{2}+1+1=590$$

Add 1 and 1 to get 2.

$$3x^{2}+2=590$$

Subtract 590 from both sides.

$$3x^{2}+2-590=0$$

Subtract 590 from 2 to get -588.

$$3x^{2}-588=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 0 for b, and -588 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\times 3\left(-588\right)}}{2\times 3}$$

Square 0.

$$x=\frac{0±\sqrt{-4\times 3\left(-588\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{0±\sqrt{-12\left(-588\right)}}{2\times 3}$$

Multiply -12 times -588.

$$x=\frac{0±\sqrt{7056}}{2\times 3}$$

Take the square root of 7056.

$$x=\frac{0±84}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{0±84}{6}$$

Now solve the equation x=\frac{0±84}{6} when ± is plus. Divide 84 by 6.

$$x=14$$

Now solve the equation x=\frac{0±84}{6} when ± is minus. Divide -84 by 6.

$$x=-14$$

The equation is now solved.

$$x=14 x=-14$$