Find the sum of all possible values of $a$ such that the following equation $(x - a)^2 + (x^2 - 3x + 2)^2 = 0$ has real root in $x$ :-

Question

Find the sum of all possible values of $a$ such that the following equation $(x - a)^2 + (x^2 - 3x + 2)^2 = 0$ has real root in $x$ :-

What I Tried :- I know $(x^2 - 3x + 2) = (x - 1)(x - 2)$ .

So :- $(x - a)^2 = - (x^2 - 3x + 2)^2$

=> $(x - a) = -(x^2 - 3x + 2)$

=> $(x - a) = -(x - 1)(x - 2) = (1 - x)(x - 2)$ .

From here I don't have a good hint or a clue to move forward . Can anyone help ?

Answer 1

Hint:sum of two non negative terms can be zero if they are both zero.

Answer 2

Just note that $$ (x - a)^2 + (x - 3x + 2)^2 = 0 $$ $$ \implies (x -a) = (x^2 - 3x + 2) = (x - 2)(x - 1) = 0 \quad [\text{As }x, a \in \mathbb{R}] $$ Then $a = 2, a = 1$. Then $$ {\text{Sum of all possible valus of } a} = 2 + 1 = 3 $$