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
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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
- 2
- 3
- 4
Option: 3
Explanation:
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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