Solve -1/2=left(a-2right)^2-left(a+2right)^2+a^2/2left(a-2right)left(aright)

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

Solve for a

$$a=5$$

Steps Using Factoring

Variable a cannot be equal to any of the values 0,2 since division by zero is not defined. Multiply both sides of the equation by 2a\left(a-2\right), the least common multiple of 2,2\left(a-2\right)a.

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

Use the distributive property to multiply -a by a-2.

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

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

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

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

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

To find the opposite of a^{2}+4a+4, find the opposite of each term.

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

Combine a^{2} and -a^{2} to get 0.

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

Combine -4a and -4a to get -8a.

$$-a^{2}+2a=-8a+4-4+a^{2}$$

Subtract 4 from 4 to get 0.

$$-a^{2}+2a=-8a+a^{2}$$

Add 8a to both sides.

$$-a^{2}+2a+8a=a^{2}$$

Combine 2a and 8a to get 10a.

$$-a^{2}+10a=a^{2}$$

Subtract a^{2} from both sides.

$$-a^{2}+10a-a^{2}=0$$

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

$$-2a^{2}+10a=0$$

Factor out a.

$$a\left(-2a+10\right)=0$$

To find equation solutions, solve a=0 and -2a+10=0.

$$a=0 a=5$$

Variable a cannot be equal to 0.

$$a=5$$

Steps Using the Quadratic Formula

Variable a cannot be equal to any of the values 0,2 since division by zero is not defined. Multiply both sides of the equation by 2a\left(a-2\right), the least common multiple of 2,2\left(a-2\right)a.

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

Use the distributive property to multiply -a by a-2.

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

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

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

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

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

To find the opposite of a^{2}+4a+4, find the opposite of each term.

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

Combine a^{2} and -a^{2} to get 0.

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

Combine -4a and -4a to get -8a.

$$-a^{2}+2a=-8a+4-4+a^{2}$$

Subtract 4 from 4 to get 0.

$$-a^{2}+2a=-8a+a^{2}$$

Add 8a to both sides.

$$-a^{2}+2a+8a=a^{2}$$

Combine 2a and 8a to get 10a.

$$-a^{2}+10a=a^{2}$$

Subtract a^{2} from both sides.

$$-a^{2}+10a-a^{2}=0$$

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

$$-2a^{2}+10a=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, 10 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a=\frac{-10±\sqrt{10^{2}}}{2\left(-2\right)}$$

Take the square root of 10^{2}.

$$a=\frac{-10±10}{2\left(-2\right)}$$

Multiply 2 times -2.

$$a=\frac{-10±10}{-4}$$

Now solve the equation a=\frac{-10±10}{-4} when ± is plus. Add -10 to 10.

$$a=\frac{0}{-4}$$

Divide 0 by -4.

$$a=0$$

Now solve the equation a=\frac{-10±10}{-4} when ± is minus. Subtract 10 from -10.

$$a=-\frac{20}{-4}$$

Divide -20 by -4.

$$a=5$$

The equation is now solved.

$$a=0 a=5$$

Variable a cannot be equal to 0.

$$a=5$$

Steps for Completing the Square

Variable a cannot be equal to any of the values 0,2 since division by zero is not defined. Multiply both sides of the equation by 2a\left(a-2\right), the least common multiple of 2,2\left(a-2\right)a.

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

Use the distributive property to multiply -a by a-2.

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

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

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

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

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

To find the opposite of a^{2}+4a+4, find the opposite of each term.

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

Combine a^{2} and -a^{2} to get 0.

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

Combine -4a and -4a to get -8a.

$$-a^{2}+2a=-8a+4-4+a^{2}$$

Subtract 4 from 4 to get 0.

$$-a^{2}+2a=-8a+a^{2}$$

Add 8a to both sides.

$$-a^{2}+2a+8a=a^{2}$$

Combine 2a and 8a to get 10a.

$$-a^{2}+10a=a^{2}$$

Subtract a^{2} from both sides.

$$-a^{2}+10a-a^{2}=0$$

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

$$-2a^{2}+10a=0$$

Divide both sides by -2.

$$\frac{-2a^{2}+10a}{-2}=\frac{0}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$a^{2}+\frac{10}{-2}a=\frac{0}{-2}$$

Divide 10 by -2.

$$a^{2}-5a=\frac{0}{-2}$$

Divide 0 by -2.

$$a^{2}-5a=0$$

Divide -5, the coefficient of the x term, by 2 to get -\frac{5}{2}. Then add the square of -\frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$a^{2}-5a+\left(-\frac{5}{2}\right)^{2}=\left(-\frac{5}{2}\right)^{2}$$

Square -\frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$a^{2}-5a+\frac{25}{4}=\frac{25}{4}$$

Factor a^{2}-5a+\frac{25}{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}.

$$\left(a-\frac{5}{2}\right)^{2}=\frac{25}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(a-\frac{5}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Simplify.

$$a-\frac{5}{2}=\frac{5}{2} a-\frac{5}{2}=-\frac{5}{2}$$

Add \frac{5}{2} to both sides of the equation.

$$a=5 a=0$$

Variable a cannot be equal to 0.

$$a=5$$