SOLUTION: The roots of the equation x² + 2px + q = 0 differ by 2. Show that p² = 1 + q

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Question 1132103: The roots of the equation x² + 2px + q = 0 differ by 2. Show that
p² = 1 + q
Found 3 solutions by MathLover1, Boreal, greenestamps:
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

The roots of the equation x%5E2+%2B+2px+%2B+q+=+0 differ by+2. Show that
p%5E2+=+1+%2B+q+
x%5E2+%2B+2px+%2B+q+=+0
Comparing with ax%5E2+%2B+bx+%2B+c+=+0 we have a+=+1, b+=+2p, c+=+q
Let alpha and beta be the roots of given quadratic equation.
alpha+-beta++=+2 ......(i) [Given]
alpha%2B+beta+=+-b%2Fa+=+-2p%2F1+=+-2p
Also, alpha%2Abeta=+c%2Fa+=+q%2F1+=+q
We know that,
%28alpha+-beta+%29%5E2+=+%28alpha%2Bbeta%29%5E2++-4alpha%2Abeta..substitute values above
+%282%29%5E2+=+%28-2p%29%5E2+-4+%28q%29
4+=+4p%5E2+-4q...simplify
1+=+p%5E2+-q
+p%5E2+=+1%2B+q
Hence proved.


Answer by Boreal(15235) About Me  (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) About Me  (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: -2p+=+r%2B%28r%2B2%29+=+2r%2B2 --> -p+=+r%2B1 and
(2) the product of the roots is q: q+=+r%28r%2B2%29+=+r%5E2%2B2r

Then

p%5E2+=+%28r%2B1%29%5E2+=+r%5E2%2B2r%2B1 and
q%2B1+=+r%5E2%2B2r%2B1

And so p^2 = q+1.