Find the range of values of k for which x²+kx-3k<5 for some x, i.e. the curve y=x²+kx-3k goes below y=5

We know that x²+kx-3k-5<0 for some x for the values the k that we are trying to find.This will only occur when the curve has two distinct intersections with the x-axis. There are two distinct intersections only when the discriminant of the quadratic equation is more than 0. So k²-4(1)(-3k-5)>0 which is simplified as k²+12k+20>0.To find the values of k which satisfy this we solve k²+12k+20=0 by factorising and getting (k+10)(k+2)=0 so k=-10 or k=-2.Using these intersections with the x-axis we can sketch the graph of this quadratic and see that it is greater than 0 when k<-10 or k>-2. Therefore the solution to the question is k<-10 or k>-2.

PS
Answered by Peter S. Maths tutor

3661 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Integrate ∫sin²xcosxdx


What is the chain rule and how is it used?


The line AB has equation 5x+3y+3=0. It is parallel to a line with equation y=mx+7. What is m?


The number of bacteria present in a culture at time t hours is modeled by the continuous variable N and the relationship N = 2000e^kt, where k is a constant. Given that when t = 3, N = 18 000, find (a) the value of k to 3 significant figures


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