Question

If the roots of the equation x^2+2px+q=0 differ by 2; show that p^2=1+q

Answer 1

Answer:

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Answer 2

Answer:

p

2

=4q+1

Given equation x

2

+px+q=0

Let the roots be α,β

α+β=−p and αβ=q

Now, (α−β)

2

=(α−β)

2

−4αβ

⇒(α−β)

2

=p

2

−4q

Given ∣α−β| =1

(α−β)

2

=1

p

2

−4q=1