Question

I am not able to prove the second part of the question , i proved the first part

Answer 1

If the roots of the equation
\[\huge \frac{ 1 }{ (x+p) } + \frac{ 1 }{ (x+q) } = \frac{ 1 }{ r }\]
are equal in Magnitude but opposite in sign , show that :-
(a)p+q = 2r
and the
(b) product of the roots is equal to (-1/2)(p^2 + q^2)

Answer 2

@Kainui

Answer 3

How did you get the first part? If you get that, the second should rather be staring at you.

Did you manage \(x^{2} + (p+q-2r)x + (qp-rp-rq) = 0\) as an equivalent equation?

Answer 4

\[\huge x ^{2} + (p+q-2r)x -[\frac{ \left( p+q)r- pq \right) }{ r }]\]
Let (p+q)r - pq be k (a constant)
x^2 + (p+q-2r)x - k = 0

If one root is a and the other b

a+b=0<-------

Henceforth on substituting a and b in the equation
and by quadratic formula finding out a and b
then a+b = -p - q +2r

0 = -p -q +2r
p+q=2r

Answer 5

QED

Answer 6

Since the solutions are equal and opposite, this must represent a Difference of Squares and has to look like this. \(x^{2} - a^{2} = 0\)

Thus, \(p + q - 2r = 0\;and\;p+q = 2r\)

Thus, ... Hold on. How did you get 'r' in the denominator in that constant term?

Answer 7

I simmplified the main equaation , ( i used wolfram alpha)
and yes the constant is diided and also multiplied by r so it kindoff cancels out

Answer 8

Kinda, but not. You should have only \(-(r(p+q) - pq)\). Something went wrong in there. There shoudl be no 'r' in the denominator. That denominator should be unity (1).

Answer 9

Yes that's what it is 1 after i simplify it further i missed out r earlier

Answer 10

If you agree the constant term is -(r(p+q) - pq), then we're almost done.

Answer 11

It is -[(p+q)r -pq]) that 's what i got oh yes u got the same

Answer 12

From the result in Part 1, we have \(r = \dfrac{p+q}{2}\). Substitute and simplify. See what happens.

Answer 13

what to substitute ? and how does that make Product of roots

Answer 14

@tkhunny r u there

Answer 15

@tkhunny @tkhunny @tkhunny @tkhunny @tkhunny @tkhunny
R U there!

Answer 16

Are you forgetting your algebra?

Roots: a and b.
Factors: (x-a)(x-b)
Equation: \(x^{2} -(a+b)x + ab = 0\)

See how the coefficient on the x-term is the sum of the roots and the constant term is the product of the roots? That is what we need to see.

You have r(p+q) + pq and you have r = (p+q)/2. Substitute the second into he first and simplify things.

Answer 17

...and, never do that again. :-(