Question

The complete set of values of 'a' such that $x^2+ax+a^2+6a<0\forall x\in [-1,1]$ is:

• A. $(\frac{-5-\sqrt{21}}{2},\frac{-7+\sqrt{45}}{2})$
• B. None of these
• C. $(\frac{-7-\sqrt{45}}{2},\frac{-5-\sqrt{21}}{2})$
• D. $(\frac{-5+\sqrt{21}}{2},\frac{-7+\sqrt{45}}{2})$

Answer 1

The correct option is B

None of these

$f\left(x\right)=x^2+ax+a^2+6a<0$

$\Rightarrow f(-1)<0,f\left(1\right)<0,D<0$.

$\Rightarrow a^2-4(a^2+6a)>0,a^2+7a+1<0,a^2+5a+1<0$

$-8<a<0,\frac{-7-\sqrt{45}}{2}<a<\frac{-7+\sqrt{45}}{2}$
and $\frac{-5-\sqrt{21}}{2}<a<\frac{-5+\sqrt{21}}{2}$

$\Rightarrow \frac{-5-\sqrt{21}}{2}<a<\frac{-5+\sqrt{21}}{2}$