1. Let's hit this bad boy with the quadratic formula
The only thing that would make x irrational is that square root. So let's focus on that.
`sqrt(4+4(m^2-1)) = sqrt(4(1+m^2-1))=sqrt(4m^2)=2m`
So the roots are not only rational multiples of m, but they are specifically m-1 and -m-1.
Let's use the quadratic formula, eh?
`x = (-2q(p+r)+-sqrt(4q^2(p+r)^2-4(p^2+q^2)(q^2+r^2)))/(2(p^2+q^2))`
We're told the function has real roots. So the value under the square root must be positive (or zero). So...
Cancel the 4s and multiply this stuff out
Subtract like quantities from both sides
A square number cannot be less than zero, so it must be equal to zero.
And I think you can take it from there.