Question

If the range of real values of $b$ for which the equation $\left(x^4+4 x^2+4\right)-(b+4)\left(x^4+6 x^2+8\right)-(b+5)\left(x^4+8 x^2+16\right)=0$ has atleast one real solution is $[\alpha, \beta)$ then find the value of $(2 \alpha-5 \beta)$.

Answer 1

We have
$\left(x^2+2\right)^2-(b+4)\left(x^2+2\right)\left(x^2+4\right)-(b+5)\left(x^2+4\right)^2=0$
dividing by $\left( x ^2+ y ^2\right)^2$,
$\Rightarrow\left(\frac{x^2+2}{x^2+4}\right)^2-(b+4)\left(\frac{x^2+2}{x^2+4}\right)-(b+5)=0$
Let $\frac{x^2+2}{x^2+4}=t$, so we get
$t^2-(b+4) t-(b+5)=0$, where $t \in\left[\frac{1}{2}, 1\right)$ ....(1)
As roots of (1) are -1 and $b+5$, so
$\frac{1}{2} \leq b +5<1 \Rightarrow \frac{-9}{2} \leq b <-4$
As $b \in\left[\frac{-9}{2},-4\right)$, so $(2 \alpha-5 \beta)=2\left(\frac{-9}{2}\right)-5(-4)=11$.