Question

Show that the roots of the equation x^2-2(m+1/m)x+3=0 are real for all(non-zero) real values of m.​

Answer 1

Answer:

Step-by-step explanation:

Given the quadratic form

x2−2(m+1m)x+3

The discriminant is

4(m+1m)2−12

The roots are real whenever the discriminant is non-negative. So the question is, what is the smallest possible value for  (m+1/m)2 ? It turns out that it is  4  (which happens when  m=±1;  see plot below) so the minimum possible value for the discriminant is  4×4−12=4.  Since the discriminant can’t be negative, the roots are always real.

Plot of  f(m)=(m+1m)2.  By solving  f′(m)=0,  it’s not hard to show that the minimums occur at exactly  ±1.