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 $D>0\Rightarrow a^2-4\times 1\times \left(a^2+6a\right)>0$
$\Rightarrow -3a^2-24a>0\Rightarrow a^2+8a<0\Rightarrow a\in \left(-8,0\right)$

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 $-21$