Question

Show that the roots of the equation `x^2+px-q^2=0` are real for all real values of p and q.

Answer 1

Given: 

`x^2+px-q^2=0` 

Here, 

`a=1,b=p and c=-q^2` 

Discriminant D is given by: 

`D=(b^2-4ac) `

`=p^2-4xx1xx(-q^2)` 

`=(p^2+4q^2)>0`

D > 0 for all real values of p and q.
Thus, the roots of the equation are real.