Given that 2-3i is a root to the equation z^3+pz^2+qz-13p=0, show that p=-2 and q=5.

Substitute 2-3i into equation using part i (2-3i)3=-46-9i.  -46-9i+p(-5-12i)+q(2-3i)-13p=0. -46-18p+2q-9i-12pi-3iq=0. Real: -46-18p+2q=0 and Imaginary: -9-12p-3q=0. p=-2, q=5

WN
Answered by William N. Maths tutor

10637 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

What are logarithms and how do you manipulate them?


How would you determine what sort of stationary point this curve has? x^3 - 6x^2 + 9x - 4


A curve, C, has equation y =(2x-3)^5. A point, P, lies on C at (w,-32). Find the value of w and the equation of the tangent of C at point, P in the form y =mx+c.


A smooth 4g marble is held at rest on a smooth plane which is fixed at 30 degrees to a horizontal table. The marble is released from rest - what speed is the marble travelling at 5 seconds after being released? Let g = 9.8ms^2


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences