Question

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

Answer 1

Answer: 11

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
$\frac{1}{2}\le b+5<1\Rightarrow \frac{-9}{2}\le b<-4$
As $b\in \left[\frac{-9}{2},-4\right)$, so $(2\alpha -5\beta )=2\left(\frac{-9}{2}\right)-5(-4)=11$.