Question

Nature of roots

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 :-
p+q = 2r and the product of the roots is equal to (-1/2)(p^2 + q^2)

Answer 2

I tried simplifying it

Answer 3

After solving the above equation :

\(x^2 + x(p+q-2) + (pq-p-q) = 0\)

Answer 4

So, the roots of the above equation are equal but opposite in magnitude.

Answer 5

Yeah simplified it , i gott that ,
p+q =2r how to prove that

Answer 6

Hmm, working on it :-)

Answer 7

o.O , I did a mistake there..

Answer 8

I knew It simplifies
to
x^2 + (p+q-2r)x - [(p +q -(pq)/(r)]r = 0

Answer 9

Yeah... I didn't include "r" in that.

Answer 10

\[\huge x ^{2} + (p+q -2r)x -(p+q - \frac{ pq }{ r })r = 0\]

Answer 11

Let the constant term be : (p+q)r - pq

Answer 12

Okaay

Answer 13

As the roots are equal in magnitude and opposite in sign,

we can say that :

\(\alpha + \beta = 0 \)

I got alpha = \(\cfrac{-(p+q-2r) + \sqrt{(p+q)^2 + 4r^2}}{2}\)

and beta = \(\cfrac{-(p+q-2r) - \sqrt{(p+q)^2 + 4r^2}}{2}\)

So, \(\alpha + \beta = -(p+q-2r) = 0 \)

Therefore, -(p+q) = -2r

or :

\(\boxed{\bf{p+q = 2r}}\)

PROVED!! (1st part)

Answer 14

So , i can calculate the product of the roots now thanks!!

Answer 15

Yeah! :-) And you're welcome. Good Luck.

(It was a nice question... Thanks)

Answer 16

Yup