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

Question

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

Accepted Answer

$$\alpha  + \beta  = a - 3,\,\alpha \beta  = 1 - 2a$$

$$ \Rightarrow {\alpha ^2} + {\beta ^2} = {(a - 3)^2} - 2(1 - 2a)$$

$= {a^2} - 6a + 9 - 2 + 4a$

$= {a^2} - 2a + 7$

$= {(a - 1)^2} + 6$

So, $${\alpha ^2} + {\beta ^2} \ge 6$$

Solutions · 1