SOLUTION: The cubic equation x^3-12x^2+ax-48=0 has roots p,2p and 3p. Find the value of p.

Question 1099436: The cubic equation x^3-12x^2+ax-48=0 has roots p,2p and 3p. Find the value of p.

Answer by ikleyn(52778) (Show Source): You can put this solution on YOUR website!
.

According to Vieta's theorem, the product of the roots is equal to the constant term taken with the opposite sign in this case.


In the mathematical form,  p*(2p)*(3p) = 48,   or

6p^3 = 48  ====>  p^3 =  = 8  ====>  p =  = 2.


Answer.  p = 2.

It is INTERESTING that under the given condition, the answer DOES NO DEPEND on the value of the coefficient "a" of the given equation.

    In opposite, by knowing all the roots p= 2, 2p = 4  and  3p= 6, we can calculate the coefficient "a" 

    (using again the Vieta's theorem) as the sum of all pair-wise products of the roots:


    a = 2*4 + 2*6 + 4*6 = 8 + 12 + 24 = 44.

RELATED QUESTIONS
How would I solve: Find the values for p for which the equation; (p+3)x^2-12x+2p=0... (answered by mananth)

if P^3+3P^2+3P=7, then the value of P^2+2P... (answered by josgarithmetic)

Given that the equation x(x-2p)=q(x-p) has real roots for all real values of p and q. If... (answered by josgarithmetic,MathTherapy)

Find the value of p for which the equation (1-2p)x^2 + 8px - (2+8p) = 0 has two equal... (answered by robertb,MathTherapy)

the first numbers p-1, 2p-2 and 3p-1 are the first three terms of a GP where p>0 find... (answered by rothauserc)

The cubic equation x^3+ax^2+bx+c=0 has roots a^2,B^2 and y^2. Show that c=-4/9 and find... (answered by ikleyn)

The roots of (x^3) + 2p(x^2) - px + 10 = 0 are integral and form an arithmetic sequence.... (answered by ikleyn,greenestamps)

The equation {{{x^2-3x-2=0}}} has roots {{{alpha}}} and {{{beta}}}, and the equation... (answered by Edwin McCravy)

Given that one of the roots of the quadratic equation 4x^2-(p-2)x-2p=5 is negative of the (answered by stanbon)