Answers>Maths>IB>Article

The quadratic equation x^2 - 2kx + (k - 1) = 0 has roots α and β such that α^2 + β^2 = 4. Without solving the equation, find the possible values of the real number k.

We know in a quadratic x^2 +bx + c = 0, -b/a = α + β and c/a = αβ. 

Therefore, α + β = -(-2k) = 2k, and αβ = k - 1. (Both are divided by the coefficient in front of x which is 1 so can be ignored.

Now (α + β)^2 = α^2 + 2αβ + β^2

Rearranging: α^2 + β^2 = (α + β)^2 - 2αβ

Substituting: 4 = (2k)^2 - 2(k - 1)

Expand: 4 = 4k^2 - 2k + 2 = 2 (2k^2 - k + 1)

Put all on one side: 0 = 2k^2 - k - 1 = (2k + 1)(k - 1)

Hence k = 1  or -1/2

RT
Answered by Ralph T. Maths tutor

16200 Views

See similar Maths IB tutors

Related Maths IB answers

All answers ▸

Find the coordinates that correspond to the maximum point of the following equation: y = −16x^2 + 160x - 256


Differentiate, from first principles, y=x^2


Consider the functions f and g where f(x)=3x-5 and g(x)=x-2. (a) Find the inverse function for f. (b) Given that the inverse of g is x+2, find (g-1 o f)(x).


The sum of the first n terms of an arithmetic sequence is Sn=3n^2 - 2n. How can you find the formula for the nth term un in terms of n?


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