SOLUTION: If one root of 5x^2+13x+p=0 be reciprocal of the other then the value of p is

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Question 966986: If one root of 5x^2+13x+p=0 be reciprocal of the other then the value of p is
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
If the roots of 5x%5E2%2B13x%2Bp=0 are r and 1%2Fr ,
then we can "factorize" 5x%5E2%2B13x%2Bp like this
5x%5E2%2B13x%2Bp=5%28x-r%29%28x-1%2Fr%29 ,
and if we multiply the two expressions in brackets, we get
.
So, since that is true for all values of x,
the coefficients must be the equal, meaning that
highlight%28p=5%29 ,
and 13=-5%281%2Fr%2Br%29 .
We can find those reciprocal roots,
or at least prove that there are real roots,
to make sure we have not been tricked.
It turns out that the determinant for 5x%5E2%2B13x%2B5=0 is
13%5E3-4%2A5%2A5=169-100%3E0 and there are real roots given by
x=%28-13+%2B-+sqrt%2869%29%29%2F10 .