Given the equation x^3-12x^2+ax-48=0 has roots p, 2p and 3p, find p and a.

QUESTION: Given the equation x3-12x2+ax-48 = 0 has roots p, 2p and 3p, find p and a. Roots mean x = p, x = 2p and x = 3p hence (x-p), (x-2p) and (x-3p) are factors of the equation. Expanding these three factors together will equal the equation. (x-p)(x-2p)(x-3p) = (x2-px-2px+2p2)(x-3p) = (x3-px2-2px2+2p2x-3px2+3p2x+6p2x-6p3) = 0. By collecting and equating coefficients both p and a can be found. x3+(-p-2p-3p)x2+(2p2+3p2+6p2)x-6p3 = x3-6px2+11p2x-6p3 -6p3 = -48 hence p3 = 8 and so p = 2 11p2 = a and so a = 44 ANSWER: p = 2 and a = 44

Related Further Mathematics A Level answers

All answers ▸

Find the solution the the differential equation d^2y/dx^2 + (3/2)dy/dx + y = 22e^(-4x)


How do I find the square root of a complex number?


Express cos(4x) in terms of powers of cos(x)


solve the equation 4cos^2(x) -15sin(x) = 13


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences