Solve (2a-5)^2/2-(a-3)^2+1/2geqa^2

$$\frac { ( 2 a - 5 ) ^ { 2 } } { 2 } - ( a - 3 ) ^ { 2 } + \frac { 1 } { 2 } \geq a ^ { 2 }$$

Solve for a

$$a\leq 1$$

Solution Steps

Multiply both sides of the equation by 2. Since 2 is positive, the inequality direction remains the same.

$$2\left(\frac{\left(2a-5\right)^{2}}{2}-\left(a-3\right)^{2}\right)+1\geq 2a^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2a-5\right)^{2}.

$$2\left(\frac{4a^{2}-20a+25}{2}-\left(a-3\right)^{2}\right)+1\geq 2a^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(a-3\right)^{2}.

$$2\left(\frac{4a^{2}-20a+25}{2}-\left(a^{2}-6a+9\right)\right)+1\geq 2a^{2}$$

To find the opposite of a^{2}-6a+9, find the opposite of each term.

$$2\left(\frac{4a^{2}-20a+25}{2}-a^{2}+6a-9\right)+1\geq 2a^{2}$$

Use the distributive property to multiply 2 by \frac{4a^{2}-20a+25}{2}-a^{2}+6a-9.

$$2\times \frac{4a^{2}-20a+25}{2}-2a^{2}+12a-18+1\geq 2a^{2}$$

Express 2\times \frac{4a^{2}-20a+25}{2} as a single fraction.

$$\frac{2\left(4a^{2}-20a+25\right)}{2}-2a^{2}+12a-18+1\geq 2a^{2}$$

Cancel out 2 and 2.

$$4a^{2}-20a+25-2a^{2}+12a-18+1\geq 2a^{2}$$

Combine 4a^{2} and -2a^{2} to get 2a^{2}.

$$2a^{2}-20a+25+12a-18+1\geq 2a^{2}$$

Combine -20a and 12a to get -8a.

$$2a^{2}-8a+25-18+1\geq 2a^{2}$$

Subtract 18 from 25 to get 7.

$$2a^{2}-8a+7+1\geq 2a^{2}$$

Add 7 and 1 to get 8.

$$2a^{2}-8a+8\geq 2a^{2}$$

Subtract 2a^{2} from both sides.

$$2a^{2}-8a+8-2a^{2}\geq 0$$

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

$$-8a+8\geq 0$$

Subtract 8 from both sides. Anything subtracted from zero gives its negation.

$$-8a\geq -8$$

Divide both sides by -8. Since -8 is negative, the inequality direction is changed.

$$a\leq \frac{-8}{-8}$$

Divide -8 by -8 to get 1.

$$a\leq 1$$