Solve I=(a-5/2)^2+a^2

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

Solve for I

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

Short Solution Steps

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

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

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

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

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.

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

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

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

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

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

Calculate 2 to the power of 2 and get 4.

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

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

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

Solve for a (complex solution)

$$a=-\frac{2\sqrt{5I-25}}{5}+1$$

$$a=\frac{2\sqrt{5I-25}}{5}+1$$

Solve for a

$$a=-\frac{2\sqrt{5I-25}}{5}+1$$

$$a=\frac{2\sqrt{5I-25}}{5}+1\text{, }I\geq 5$$