Solve 5={left(a-5/2right)}^2+a^2

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

Solve for a

$$a=1$$

Steps Using the Quadratic Formula

To raise \frac{a-5}{2} to a power, raise both numerator and denominator to the power and then divide.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply a^{2} times \frac{2^{2}}{2^{2}}.

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

Since \frac{\left(a-5\right)^{2}}{2^{2}} and \frac{a^{2}\times 2^{2}}{2^{2}} have the same denominator, add them by adding their numerators.

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

Do the multiplications in \left(a-5\right)^{2}+a^{2}\times 2^{2}.

$$5=\frac{a^{2}-10a+25+4a^{2}}{2^{2}}$$

Combine like terms in a^{2}-10a+25+4a^{2}.

$$5=\frac{5a^{2}-10a+25}{2^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$5=\frac{5a^{2}-10a+25}{4}$$

Divide each term of 5a^{2}-10a+25 by 4 to get \frac{5}{4}a^{2}-\frac{5}{2}a+\frac{25}{4}.

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

Swap sides so that all variable terms are on the left hand side.

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

Subtract 5 from both sides.

$$\frac{5}{4}a^{2}-\frac{5}{2}a+\frac{25}{4}-5=0$$

Subtract 5 from \frac{25}{4} to get \frac{5}{4}.

$$\frac{5}{4}a^{2}-\frac{5}{2}a+\frac{5}{4}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{5}{4} for a, -\frac{5}{2} for b, and \frac{5}{4} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

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

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

Multiply -4 times \frac{5}{4}.

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

Multiply -5 times \frac{5}{4}.

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

Add \frac{25}{4} to -\frac{25}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$a=\frac{-\left(-\frac{5}{2}\right)±\sqrt{0}}{2\times \frac{5}{4}}$$

Take the square root of 0.

$$a=-\frac{-\frac{5}{2}}{2\times \frac{5}{4}}$$

The opposite of -\frac{5}{2} is \frac{5}{2}.

$$a=\frac{\frac{5}{2}}{2\times \frac{5}{4}}$$

Multiply 2 times \frac{5}{4}.

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

Divide \frac{5}{2} by \frac{5}{2} by multiplying \frac{5}{2} by the reciprocal of \frac{5}{2}.

$$a=1$$

Steps for Completing the Square

To raise \frac{a-5}{2} to a power, raise both numerator and denominator to the power and then divide.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply a^{2} times \frac{2^{2}}{2^{2}}.

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

Since \frac{\left(a-5\right)^{2}}{2^{2}} and \frac{a^{2}\times 2^{2}}{2^{2}} have the same denominator, add them by adding their numerators.

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

Do the multiplications in \left(a-5\right)^{2}+a^{2}\times 2^{2}.

$$5=\frac{a^{2}-10a+25+4a^{2}}{2^{2}}$$

Combine like terms in a^{2}-10a+25+4a^{2}.

$$5=\frac{5a^{2}-10a+25}{2^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$5=\frac{5a^{2}-10a+25}{4}$$

Divide each term of 5a^{2}-10a+25 by 4 to get \frac{5}{4}a^{2}-\frac{5}{2}a+\frac{25}{4}.

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

Swap sides so that all variable terms are on the left hand side.

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

Subtract \frac{25}{4} from both sides.

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

Subtract \frac{25}{4} from 5 to get -\frac{5}{4}.

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

Divide both sides of the equation by \frac{5}{4}, which is the same as multiplying both sides by the reciprocal of the fraction.

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

Dividing by \frac{5}{4} undoes the multiplication by \frac{5}{4}.

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

Divide -\frac{5}{2} by \frac{5}{4} by multiplying -\frac{5}{2} by the reciprocal of \frac{5}{4}.

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

Divide -\frac{5}{4} by \frac{5}{4} by multiplying -\frac{5}{4} by the reciprocal of \frac{5}{4}.

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

Divide -2, the coefficient of the x term, by 2 to get -1. Then add the square of -1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Add -1 to 1.

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

Factor a^{2}-2a+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(a-1\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(a-1\right)^{2}}=\sqrt{0}$$

Simplify.

$$a-1=0 a-1=0$$

Add 1 to both sides of the equation.

$$a=1 a=1$$

The equation is now solved. Solutions are the same.

$$a=1$$