Question

The sum of the roots of \[{1 \over {x+a}}+{1 \over {x+b}}={1 \over c}\] is Zero. The products of the roots is:-

Answer 1

\[Ans.-{{-1\over2}(a^2+b^2)}\]

Answer 2

I think firstly you should solve to make it quadratic equation..

Answer 3

I wants full solution.

Answer 4

Take the LCM of the denominator...

Answer 5

What is the LCM or LCD of (x+a) and (x+b) ??

Answer 6

I gt\[{(x+b)+(x+a)}\over{(x+a)(x+b)}\]

Answer 7

Yes you are right..

Now cross multiply...

Answer 8

\[x^2+xb+ax+ab\]

Answer 9

You have denominator too..

Now cross multiply:

\[\frac{a}{b} = \frac{c}{d}\]

\[a \times d = b \times c\]

Answer 10

\[c(x+a+x+b)=(x^2+xb+ax+ab)\]

Answer 11

M i right???

Answer 12

Distribute c inside the parenthesis..

Answer 13

Have faith on yourself..
You are going right..

Answer 14

\[2cx+ca+cb=x^2+xb+ax+ab\]

Answer 15

Now can you take all the terms to one side only..

take all the Left hand side terms to right hand side..

Remember their sign will get reverse..

Answer 16

\[2cx+ca+cb-x^2-xb-ax-ab=0\]

Answer 17

You done opposite to that..
Does not matter..

Multiply (-1) both the sides..

Answer 18

\[-1(2cx+ca+cb-x^2-xb-ax-ab)=-1(0)\]

Answer 19

Apply -1 to the brackets..

Answer 20

\[-2cx-ca-cb+x^2+xb+ax+ab=0\]

Answer 21

Now from all the terms containing x take x common..
Able to do that??

Answer 22

I could'nt understand wht u wants to say:(

Answer 23

What are the x terms there?
Tell me the entire terms that contain x.

Answer 24

-2cx,x^2,xb,ax

Answer 25

I am saying the terms which contains x only not x squared and not constants..

tell me the terms having x..

Answer 26

2cx, xb ,ax

Answer 27

Now in these terms x is common and present in all the terms..
Take x common from it..

Getting??

Answer 28

x(2c),x(b),x(a)

Answer 29

From all simultaneously..

Like if I have:

x + 7xb + 8xa = x(1+ 7b + 8a)

Like this..

Answer 30

Is the question answered or should i enter in between ?

Answer 31

Ya sure you can give your opinion too @mathslover

Answer 32

Ok i will like to give wait

Answer 33

Please zoom to see it as a good file :) .. i got this and m solving further

Answer 34

@waterineyes any suggestion ?

Answer 35

Very Good..

Answer 36

But what next ?

Answer 37

I am just going all around :

Answer 38

the equation looks like x^2+(a+b-2c)x +ab-ac-bc=0
but given a+b-2c=0
i.e. a+b=2c
product of roots is ab-ac-bc=ab-c(a+b)
=ab-((a+b)^)/2= 1/2 (2ab -a^2 -b^2 -2ab) =-1/2 (a^2 +b^2)....ans

Answer 39

I got it..
Wait..

Answer 40

i also gt
thnx a lot
& here the question closes;)

Answer 41

mention not..

Answer 42

\[x^2 + bx + ax + ab - ac - bc - 2cx = 0\]

\[x^2 + (a+b-2c)x + (ab -ac-bc) = 0\]


From here, A = 1, B = (a + b - 2c) and C = (ab -ac- bc)


Sum of Roots = 0

\[-(a+b -2c) = 0\]

\[c = \frac{a+b}{2}\]

Product of Roots = C

\[ab - ac - bc = ab - a(\frac{a+b}{2}) - b(\frac{a+b}{2})\]

\[= \frac{2ab - a^2 - ab - ba - b^2}{2} = \frac{-a^2 - b^2}{2} = \frac{-1}{2}(a^2 + b^2)\]

Answer 43

good work @matricked and @waterineyes

Answer 44

Thanks buddy..

Answer 45

@mathslover mention not