Question 90

# 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

Solution

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