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CAT 2017 [slot 2] Question with solution 24

Question 58:
The minimum possible value of the sum of the squares of the roots of the equation $x^2+(a+3)x-(a+5)=0$ is
  1. 1
  2. 2
  3. 3
  4. 4
Option: 3
Explanation:

Let the roots of the equation $x^2+(a+3)x-(a+5)=0 $ be equal to $p,q$

Hence, $p+q = -(a+3)$ and $p \times q = -(a+5)$

Therefore, $p^2+q^2 = a^2+6a+9+2a+10 = a^2+8a+19 = (a+4)^2+3$

As $(a+4)^2$ is always non negative, the least value of the sum of squares is 3


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