Question

Sum of the squares of all integral values of $a$ for which the inequality $x^2+ax+a^2+6a<0$ is satisfied for all $x\in \left(1,2\right)$ must be equal to

• A. $90$
• B. $89$
• C. $88$
• D. $91$

Answer 1

Answer: (D)

Let, $f\left(x\right)=x^2+ax+a^2+6a$
$\therefore$ $f(1)\le 0$

$\Rightarrow$ $a^2+7a+1<0$
or $\frac{-7-3\sqrt{5}}{2}<a<\frac{-7+3\sqrt{5}}{2}$ ....(i)
$f(2)\le 0$
$\Rightarrow$ $a^2+8a+4<0$
or $-4-2\sqrt{3}<a<-4+2\sqrt{3}$ ....(ii)
and $D>0$
$\Rightarrow a^2-4\cdot 1\left(a^2+6a\right)>0$
$\Rightarrow$ $a^2+8a<0$
or $-8<a<0$ ....(iii)
From Eqs. (i), (ii) and (iii), we get,
$\frac{-7-3\sqrt{5}}{2}\le a\le -4+2\sqrt{3}$
Hence, integral values of $a$ are $-6,-5,-4,-3,-2,-1$
Required Sum
$={\left(-6\right)}^2+{\left(-5\right)}^2+{\left(-4\right)}^2+{\left(-3\right)}^2+{\left(-2\right)}^2+{\left(-1\right)}^2$
$=91$ .