Solve a^2+(2-a)^2-2a-2(2-a)leq0

$$a ^ { 2 } + ( 2 - a ) ^ { 2 } - 2 a - 2 ( 2 - a ) \leq 0$$

Solve for a

$$a\in \begin{bmatrix}0,2\end{bmatrix}$$

Solution Steps

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

$$a^{2}+4-4a+a^{2}-2a-2\left(2-a\right)\leq 0$$

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

$$2a^{2}+4-4a-2a-2\left(2-a\right)\leq 0$$

Combine -4a and -2a to get -6a.

$$2a^{2}+4-6a-2\left(2-a\right)\leq 0$$

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

$$2a^{2}+4-6a-4+2a\leq 0$$

Subtract 4 from 4 to get 0.

$$2a^{2}-6a+2a\leq 0$$

Combine -6a and 2a to get -4a.

$$2a^{2}-4a\leq 0$$

Factor out a.

$$2a\left(a-2\right)\leq 0$$

For the product to be ≤0, one of the values a and a-2 has to be ≥0 and the other has to be ≤0. Consider the case when a\geq 0 and a-2\leq 0.

$$a\geq 0 a-2\leq 0$$

The solution satisfying both inequalities is a\in \left[0,2\right].

$$a\in \begin{bmatrix}0,2\end{bmatrix}$$

Consider the case when a\leq 0 and a-2\geq 0.

$$a-2\geq 0 a\leq 0$$

This is false for any a.

$$a\in \emptyset$$

The final solution is the union of the obtained solutions.

$$a\in \begin{bmatrix}0,2\end{bmatrix}$$