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