Solve (a-1)^2+a^2=(a+1)^2

$$( a - 1 ) ^ { 2 } + a ^ { 2 } = ( a + 1 ) ^ { 2 }$$

Solve for a

$$a=4$$

$$a=0$$

Steps Using Factoring

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

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

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

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

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

$$2a^{2}-2a+1=a^{2}+2a+1$$

Subtract a^{2} from both sides.

$$2a^{2}-2a+1-a^{2}=2a+1$$

Combine 2a^{2} and -a^{2} to get a^{2}.

$$a^{2}-2a+1=2a+1$$

Subtract 2a from both sides.

$$a^{2}-2a+1-2a=1$$

Combine -2a and -2a to get -4a.

$$a^{2}-4a+1=1$$

Subtract 1 from both sides.

$$a^{2}-4a+1-1=0$$

Subtract 1 from 1 to get 0.

$$a^{2}-4a=0$$

Factor out a.

$$a\left(a-4\right)=0$$

To find equation solutions, solve a=0 and a-4=0.

$$a=0 a=4$$

Steps Using the Quadratic Formula

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

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

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

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

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

$$2a^{2}-2a+1=a^{2}+2a+1$$

Subtract a^{2} from both sides.

$$2a^{2}-2a+1-a^{2}=2a+1$$

Combine 2a^{2} and -a^{2} to get a^{2}.

$$a^{2}-2a+1=2a+1$$

Subtract 2a from both sides.

$$a^{2}-2a+1-2a=1$$

Combine -2a and -2a to get -4a.

$$a^{2}-4a+1=1$$

Subtract 1 from both sides.

$$a^{2}-4a+1-1=0$$

Subtract 1 from 1 to get 0.

$$a^{2}-4a=0$$

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

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

Take the square root of \left(-4\right)^{2}.

$$a=\frac{-\left(-4\right)±4}{2}$$

The opposite of -4 is 4.

$$a=\frac{4±4}{2}$$

Now solve the equation a=\frac{4±4}{2} when ± is plus. Add 4 to 4.

$$a=\frac{8}{2}$$

Divide 8 by 2.

$$a=4$$

Now solve the equation a=\frac{4±4}{2} when ± is minus. Subtract 4 from 4.

$$a=\frac{0}{2}$$

Divide 0 by 2.

$$a=0$$

The equation is now solved.

$$a=4 a=0$$

Steps for Completing the Square

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

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

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

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

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

$$2a^{2}-2a+1=a^{2}+2a+1$$

Subtract a^{2} from both sides.

$$2a^{2}-2a+1-a^{2}=2a+1$$

Combine 2a^{2} and -a^{2} to get a^{2}.

$$a^{2}-2a+1=2a+1$$

Subtract 2a from both sides.

$$a^{2}-2a+1-2a=1$$

Combine -2a and -2a to get -4a.

$$a^{2}-4a+1=1$$

Subtract 1 from both sides.

$$a^{2}-4a+1-1=0$$

Subtract 1 from 1 to get 0.

$$a^{2}-4a=0$$

Divide -4, the coefficient of the x term, by 2 to get -2. Then add the square of -2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$a^{2}-4a+\left(-2\right)^{2}=\left(-2\right)^{2}$$

Square -2.

$$a^{2}-4a+4=4$$

Factor a^{2}-4a+4. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(a-2\right)^{2}=4$$

Take the square root of both sides of the equation.

$$\sqrt{\left(a-2\right)^{2}}=\sqrt{4}$$

Simplify.

$$a-2=2 a-2=-2$$

Add 2 to both sides of the equation.

$$a=4 a=0$$