JEE WhatsApp Group
Sign in
Please select an account to continue using cracku.in
↓ →
JEE WhatsApp Group
Join Our JEE Preparation Group
Prep with like-minded aspirants; Get access to free daily tests and study material.
The minimum value of the sum of the squares of the roots of $$x^2 + (3-a)x = 2(a-1)$$ is
Given equation,
$$x^2+(3-a)x=2(a-1)$$
or,
$$x^2+(3-a)x-2(a-1)=0$$
Let the roots be
$$\alpha,\beta$$
Then,
$$\alpha+\beta=a-3$$
and
$$\alpha\beta=-2(a-1)$$
Now,
$$\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta$$
$$=(a-3)^2-2[-2(a-1)]$$
$$=a^2-6a+9+4a-4$$
$$=a^2-2a+5$$
Complete the square:
$$a^2-2a+5=(a-1)^2+4$$
Since
$$(a-1)^2\ge0,$$
minimum value occurs at
$$a=1$$
Hence, minimum value is
$$\boxed{4}$$.
Click on the Email ☝️ to Watch the Video Solution
Create a FREE account and get:
Educational materials for JEE preparation
Ask our AI anything
AI can make mistakes. Please verify important information.
AI can make mistakes. Please verify important information.
