Math Olympiad question:
If the equation in $x$ has real roots, then find the value of $a$ and $b$. $x^2 + 2(1 + a)x + (3a^2 + 4ab + 4b^2 + 2) = 0$
Approach:
For at least one real root:
$$b^2 -4ac \ge 0$$
So:
$$[2(1+a)]^2 - 4(3a^2 +4ab+4b^2+2)(1)\ge 0$$
$$1+a^2+2a - 3a^2-4ab-4b^2-2\ge 0$$
$$a^2 +2ab+2b^2 -a+1\le 0$$
$$(a+b)^2 +b^2-a+1\le 0$$
What to do further? I don't think a graphical approach would work, and I'm also not able to find the value which might satisfy the above. Is there a better method?
Also I'm not sure if the last step helps.
You have an error in the line $a^2 +2ab+2b^2 -a+1\le 0$. It should be $$a^2 +2ab+2b^2 -a+\frac 1 2 \leq 0$$ and it can be rewritten as $$\frac12 (a + 2b)^2 + \frac12 (a - 1)^2 \leq 0.$$ Now you should know what to do.