Question

If the equation $x^4-a{x}^3+11x^2-ax+1=0$ has four distinct positive roots then the range of $a$ is $(m,M)$. Find the value of $m+2M$.

Answer 1

$x^4-a{x}^3+11x^2-ax+1=0$

$x^4+11x^2+1=a(x^3+x)$

$a=\frac{x^2(x^2+11+\frac{1}{x^2})}{x^2(x+\frac{1}{x})}$

$⟹a=\frac{x^2+\frac{1}{x^2}+2+9}{x+\frac{1}{x}}=\frac{{(x+\frac{1}{x})}^2+9}{x+\frac{1}{x}}$

Let $x+\frac{1}{x}=t$

$\therefore a=\frac{t^2+9}{t}$

$⟹t^2+9-at=0$

As roots are real, so $x$ is real and $\frac{1}{x}$ is also real which implies that $t$ is real.

For $t$to be distinct

$D>0$

$⟹b^2-4ac>0$

$⟹a^2-4\left(1\right)\left(9\right)>0$

$⟹a^2>36$

$⟹a<-6,a>6$

As $a$ is positive,

So, $a>6$

$m=6$

$x+\frac{1}{x}=t$

$AM\ge GM$

$⟹\frac{x+\frac{1}{x}}{2}\ge \sqrt{x\cdot \frac{1}{x}}$

$⟹x+\frac{1}{x}>2$

$\therefore t>2$

$t^2-at+9=0$

$t_1,t_2=\frac{-(-a)\pm \sqrt{(-a{)}^2-(4\times 9\times 1)}}{2}$

As, $a>2$

$\therefore a\pm \sqrt{a^2-36}>4$

$⟹a-4>\pm \sqrt{a^2-36}$

$⟹(a-4{)}^2>(\pm \sqrt{a^2-36}{)}^2$

$⟹a^2+16-8a>a^2-36$

$⟹\frac{52}{8}>a$

$\therefore a<\frac{13}{2}$

So, $M=\frac{13}{2}$

Thus, $m+2M=6+(2\times \frac{13}{2})=19$