The minimum value of the sum of the squares of the roots of x^2+(3-a)x+1=2a is:

Question

The minimum value of the sum of the squares of the roots of $x^2+(3-a)x+1=2a$ is:

Accepted Answer

Rewrite as $x^2+(3-a)x+(1-2a)=0$. Then $\alpha+\beta=a-3,\ \alpha\beta=1-2a$.

$$\alpha^2+\beta^2=(a-3)^2-2(1-2a)=a^2-6a+9-2+4a=a^2-2a+7=(a-1)^2+6\ge6,$$

with minimum $6$ at $a=1$.

Solutions · 1