Find whether roots of an equation are real or not.

Question

Is it possible to write, let's say :- (a^2 + 1/a^2 - 1) as (a-b)^2.

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So I have to solve a problem (as practice, no test or assignment) that says to show if the roots of a given equation are real.

The equation is :-

$$x^2 - 2\left(m + \frac1m \right)x + 3 = 0$$

I proceeded to solve it using the discriminant formula(b^2 - 4ac) and what I'm left with is :-

$$4\left(m^2 + \frac1{m^2} -1\right) $$

And I have no idea what to do next, i know that for a equation to be real its discriminant has to be greater than zero, but I dont know how this would be greater or less than zero and if this is the correct answer to the problem.

Thanks for the help

Answer 1

Note that the discriminant,

$$m^2 + \frac 1{m^2} -1 = \left(m - \frac 1m\right)^2+1>0$$

So, the roots are real.

Answer 2

Your discriminant should be $$\left[-2\left(m+\frac1m\right)\right]^2-4(1)(3)=4\left(m^2+2m\frac1m+\frac{1}{m^2}-3\right)=4\left(m^2-1+\frac{1}{m^2}\right)=4\left(m^2-2+\frac{1}{m^2}+1\right)=4\left(m^2-2m\frac1m+\frac{1}{m^2}+1\right)=4\left(m-\frac1m\right)^2+4,$$ which is always positive, as wanted.

Answer 3

As $x^2-2ax+3=0\iff (x-a)^2=a^2-3$ we'll have two solution iff $a^2>3$. Now $$\left(m+\frac1m\right)^2=\left(m-\frac1m\right)^2+4.$$