Question

The problem is :
if m and n be the roots of the eq. p(x^2-x)+x+5=0 If p1 and p2 are the values of p for which m and n are connected by the m/n+n/m=4/5. Find the value of (p1/p2^2) +(p2/p1^2)
now by from m/n+n/m=4/5 i get (m^2+n^2)/mn=4/5 from which I get the equation p^2-16p+1=0 by substituting the values of m+n and m.n.
The hint to the sum says show p1 and p2 are roots of the equation p^2-16p+1=0.
My question is how do i know that p1 and p2 are roots of the equation p^2-16p+1=0 which we get from m/n+n/m=4/5.

Answer 1

given m+n= 1/p -1 and mn = 5/p
so we need to eliminate m and n and should get a quadratic equation in p

Answer 2

yes i got that .. btw m+n = (p-1)/p ..so the quadratic equation i got is p^2-16p+1=0. now question is how do i know that p1 and p2 are roots of the equation p^2-16p+1=0 which we get from m/n+n/m=4/5. so that we can proceed to find the value of p1 and p2

Answer 3

no need of that
just if p1 and p2 are the roots of the quadratic eq.(as u have got) above
then p1+p2= 16 and p1*p2=1

Answer 4

but how do i know that p1 and p2 are roots of the quadratic equation we get from m/n+n/m=4/5 .. the problem says " If p1 and p2 are the values of p for which m and n are connected by the m/n+n/m=4/5." .. no where does it say that p1 and p2 will be the root of the quadratic equation we get from m/n+n/m=4/5

Answer 5

(p1/p2^2) +(p2/p1^2) =(p1^3+p2^3)/(p1^2*p2^2)
=((p1+p2)^3 - 3*p1*p2*(p1 p2))/(p1*p2)^2

Answer 6

yes i did get this part too, but my question is not this !! @primeralph could you please clarify my doubt ?

Answer 7

@primeralph : i have a doubt in this part

"but how do i know that p1 and p2 are roots of the quadratic equation we get from m/n+n/m=4/5 .. the problem says " If p1 and p2 are the values of p for which m and n are connected by the m/n+n/m=4/5." .. no where does it say that p1 and p2 will be the root of the quadratic equation we get from m/n+n/m=4/5"

Answer 8

@digitalmonk Can you write some of the stuff in math form? I can't read when there's a cluster.

Answer 9

if
p1 be a root of the p^2-16p+1=0
then
p1^2-16p1+ +1 shoul be 0

Answer 10

m and n are the roots of \[p(x^2-x)+x+5=0 \]. If p1 and p2 are the values of p for which m and n are connected by \[\frac{ m }{ n }+\frac{n}{m} = \frac{4}{5}\]. Find the value of \[(p1/p2^2) +(p2/p1^2) \] @primeralph

Answer 11

my doubt is
but how do i know that p1 and p2 are roots of the quadratic equation we get from m/n+n/m=4/5 .. the problem says " If p1 and p2 are the values of p for which m and n are connected by the m/n+n/m=4/5." .. no where does it say that p1 and p2 will be the root of the quadratic equation we get from m/n+n/m=4/5"

Answer 12

Okay, hold on.

Answer 13

It's a bunch of algebra:

Expand the original equation.
\[5+x-px + px^2=0\]Solve that for \(x\)
\[\left\{\left\{x\to \frac{-1+p-\sqrt{1-22 p+p^2}}{2 p}\right\},\left\{x\to \frac{-1+p+\sqrt{1-22 p+p^2}}{2 p}\right\}\right\}\]Now call the first one m and the second one n and plug into \[\frac{m}{n}+\frac{n}{m} = 5\]\[\frac{-1+p-\sqrt{1-22 p+p^2}}{-1+p+\sqrt{1-22 p+p^2}}+\frac{-1+p+\sqrt{1-22 p+p^2}}{-1+p-\sqrt{1-22 p+p^2}}=\frac{4}{5}\]Amazingly, that simplifies to \[\frac{1}{p}+p=16\]

Answer 14

@whpalmer4 Beat me to it.

Answer 15

Keep on going, I'm calling it a night...

Answer 16

thanks @whpalmer4 but i still don't understand how p1 and p2 are the values of p for which m and n are connected by the m/n+n/m=4/5.become the roots of the quadratic equation we get from m/n+n/m=4/5...

Answer 17

the problem says they are...