For what values of $\alpha > 0$ the equation $p(x) = x^3-9x^2+26x-\alpha =0$ has three positive real roots?

For what values of $\alpha > 0$ the equation $p(x) = x^3-9x^2+26x-\alpha =0$ has three positive real roots?

The options given are as follows :

$(A)$ $\alpha \ge 27$

$(B)$ $\alpha > 81$

$(C)$ $27 < \alpha <81$

$(D)$ $54 < \alpha \le 81$

What I have tried is that I first transform the equation to the form $y^3 + 3Hy + G = 0$ and then I have applied the fact that this equation has three positive real roots if $G^2 + 4H^3 < 0$.Here I found $G = 24 - \alpha$ and $H = -\frac {1} {3}$ and then calculating we get $\alpha \in (24-\frac {2 {\sqrt 3}} {9} , 24 + \frac {2 {\sqrt 3}} {9})$ which is not similar to any of the given options.Please help me in finding the right option.

Thank you in advance.

2 Answers

Consider the roots are $a$,$b$, $c$.

Consider $$(x-a)(x-b)(x-c)=x^3-(a+b+c)x^2+(ab+bc+ac)x-abc$$ So $$ab+bc+ac=26,\alpha=abc$$ By AM-GM inequality $$\sqrt[3]{ab\cdot bc\cdot ac}\le\frac{ab+bc+ac}{3}\implies\alpha^{2/3}\le\frac{26}3<9\implies\alpha<27$$

So no options are correct.

I plotted the graph with desmos. I think you are correct.

We can also solve the problem by calculus.

$p'(x)=3x^2-18x+26$ and $p''(x)=6x-18$.

When $x=3+\frac{1}{\sqrt{3}}$, $p'(x)=0$ and $p''(x)>0$.

When $x=3-\frac{1}{\sqrt{3}}$, $p'(x)=0$ and $p''(x)<0$.

$p(x)$ attains its maximum when $x=3-\frac{1}{\sqrt{3}}$ and its minimum when $x=3+\frac{1}{\sqrt{3}}$

$p(x)=0$ has $3$ positive roots if

(1) $p(3+\frac{1}{\sqrt{3}})<0$,

(2) $p(3-\frac{1}{\sqrt{3}})>0$, and

(3) $p(0)<0$ (this is true whenever (2) is true).