Question

If the inequality $x^{2}+ax+a^{2}+6a < 0$ is satisfied for all $x\in \left(1,2\right),$ then the sum of all the integral values of $a$ must be equal to

• A. $-10$
• B. $-15$
• C. $-21$
• D. $-28$

Answer 1

Answer: (C)

$f\left(x\right)=x^{2}+ax+a^{2}+6a$

$f\left(1\right) < 0,f\left(2\right) < 0$ and $D>0$
$f\left(1\right) < 0\Rightarrow a^{2}+7a+1 < 0\Rightarrow a\in \left(\frac{- 7 - 3 \sqrt{5}}{2} , \frac{- 7 + 3 \sqrt{5}}{2}\right)$
$f\left(2\right) < 0\Rightarrow a^{2}+8a+4 < 0\Rightarrow a\in \left(- 4 - 2 \sqrt{3} , - 4 + 2 \sqrt{3}\right)$
and $D>0\Rightarrow a^{2}-4\times 1\times \left(a^{2} + 6 a\right)>0$
$\Rightarrow -3a^{2}-24a>0\Rightarrow a^{2}+8a < 0\Rightarrow a\in \left(- 8,0\right)$

Taking intersection, we get, $a\in \left(\frac{- 7 - 3 \sqrt{5}}{2} , - 4 + 2 \sqrt{3}\right)$
Integral values of $a=-6,-5,-4,-3,-2,-1$
Hence, the sum of values is $-21$