SOLUTION: The roots of the equation x� + 2px + q = 0 differ by 2. Show that p� = 1 + q

Question 1132103: The roots of the equation x� + 2px + q = 0 differ by 2. Show that
p� = 1 + q
Found 3 solutions by MathLover1, Boreal, greenestamps:
Answer by MathLover1(20849)   (Show Source): You can put this solution on YOUR website!

The roots of the equation differ by. Show that

Comparing with we have , ,
Let and be the roots of given quadratic equation.
......(i) [Given]

Also,
We know that,
..substitute values above

...simplify

Hence proved.

Answer by Boreal(15235)   (Show Source): You can put this solution on YOUR website!
roots are (1/2)(-2+/- sqrt (4p^2-4q))
sqrt(4p^2-4q)=2*sqrt(p^2-q). 2s cancel
roots are -1+sqrt(p^2-q) and -1- sqrt(p^2-q)
the difference is 2sqrt(p^2-q) and that equals 2
so sqrt(p^2-q)=1
p^2-q=1, squaring both sides
p^2=1+q

Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!

Let the roots be r and r+2. Vieta's Theorem tells us

(1) the sum of the roots is -2p: --> and
(2) the product of the roots is q:

Then

and

And so p^2 = q+1.
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