Answer to Question #240268 in Algebra for Ahmed

Question

Equation X^3 + 21X^2 + p = 0$$ has three different real solutions that form an arithmetic progression. Find the largest value of parameter p.

Answer 1

Solution:

$x^3+21x^2+p=0$

It can be written as $(x-a)(x-b)(x-c)=0$

Here, roots are $a,b,c$.

Then, $x^3+21x^2+p=(x-a)(x-b)(x-c)$

$\Rightarrow x^3+21x^2+p=(x-a)(x^2-x(b+c)+bc) \\ \Rightarrow x^3+21x^2+p=x^3-ax^2-x^2(b+c)+ax(b+c)+xbc-abc \\ \Rightarrow x^3+21x^2+p=x^3+x^2(-a-b-c)+x(ab+ac+bc)-abc$

On comparing both sides,

$-a-b-c=21;\ ab+ac+bc=0;-abc=p \\\Rightarrow a+b+c=-21\ ...(i);\ ab+ac+bc=0;abc=-p$

It is given that the roots are in AP.

$a+c=2b\ ...(ii)$

using (ii) in (i),

$3b=-21 \\ \Rightarrow b=-7$

Put this value in (ii),

$a+c=-14$

Also, $abc=-p$

$\Rightarrow ac(-7)=-p$ [Using b=-7]

$\Rightarrow ac=\dfrac p7$

Also, $ab+ac+bc=0$

$\Rightarrow -7a+ac-7c=0$ [Using b=-7]

$\Rightarrow 7(a+c)=ac \\\Rightarrow 7(-14)=\dfrac p7 \\ \Rightarrow p=-686$