JEE MAIN - Mathematics (2022 - 26th July Evening Shift - No. 1)
The minimum value of the sum of the squares of the roots of $$x^{2}+(3-a) x+1=2 a$$ is:
4
5
6
8
Explanation
_26th_July_Evening_Shift_en_1_1.png)
$$\alpha + \beta = a - 3,\,\alpha \beta = 1 - 2a$$
$$ \Rightarrow {\alpha ^2} + {\beta ^2} = {(a - 3)^2} - 2(1 - 2a)$$
$$ = {a^2} - 6a + 9 - 2 + 4a$$
$$ = {a^2} - 2a + 7$$
$$ = {(a - 1)^2} + 6$$
So, $${\alpha ^2} + {\beta ^2} \ge 6$$
Comments (0)
