Use algebra to find the set of values of x for which mod(3x^2 - 19x + 20) < 2x + 2.

The initial quadratic can be either positive or negative so we must solve for both possibilities.

Solving for positive:

3x^2 - 19x + 20 < 2x + 2    =    3x^2 - 21x + 18 < 0

                                           =    x^2 - 7x + 6 < 0

                                           =    (x - 6)(x - 1) < 0

Therefore, Critical Values where x crosses the x-axis are x = 1 or x = 6. And since we are solving for < 0, we focus on the graph under the x-axis, resulting in 1 < x < 6.

Solving for negative:

-3x^2 + 19x - 20 < 2x + 2    =    3x^2 - 17x + 22 > 0

                                            =    (3x - 11)(x - 2) > 0

Therefore, Critical Values where x crosses the x-axis are x = 2 or x = 11/3. And since we are solving for > 0, we focus on the graph above the x-axis, resulting in x < 2 or x > 11/3.

We must then find the values for x which satisfies both positive and negative, and using a simple numberline we find 1 < x < 2 or 11/3 < x < 6 as our final answers.

JM
Answered by James M. Further Mathematics tutor

10143 Views

See similar Further Mathematics A Level tutors

Related Further Mathematics A Level answers

All answers ▸

How do I prove that the differential of coshx is equal to sinhx?


differentiate arsinh(cosx))


MEI (OCR) M4 June 2006 Q3


using an integrating factor, find the general solution of the differential equation dy/dx +y(tanx)=tan^3(x)sec(x)


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