Question

Question Based on nature of roots

Answer 1

If a ,b ,c belong to R , then prove that the roots of the equation
\[\frac{ 1 }{ x-a } + \frac{ 1 }{ x-b } + \frac{ 1 }{ x-c }=0\] are always real and cannot have roots a=b=c

Answer 2

multiply throughout by (x-a)(x-b)(x-c)

you'll get a quadratic
find its discriminant

Answer 3

Okay wait

Answer 4

http://www.wolframalpha.com/input/?i=%5B1%2F%28x-a%29%5D+%2B%5B1%2F%28x-b%29%5D+%2B+%5B1%2F%28x-c%29%5D

Answer 5

A =3
B = -2(a+b+c)
C= ab+ac+bc

right ?

Answer 6

yes

Answer 7

finding the delta

Answer 8

if we want to prove that roots are real, we have to prove
B^2>4AC

Answer 9

or
(a+b+c)^2 > 3(ab+bc+ac)

right ??

Answer 10

Yes

Answer 11

tried ?

Answer 12

so we need to prove \(
a^2+b^2+c^2-ab-bc-ca\) is always positive.

it'll be positive if its a whole squre....or sum of whole squares...

\(\begin{align}
& a^2+b^2+c^2-ab-bc-ca\geq0 \\
&\iff \frac12(2a^2+2b^2+2c^2-2ab-2bc-2ca)\geq0 \\
&\iff \frac12((a^2-2ab+b^2)+(b^2-2bc+c^2)+(c^2-2ca+a^2))\geq0\\
&\iff \frac12((a-b)^2+(b-c)^2+(c-a)^2)\geq0\\
\end{align}\)

Answer 13

Thabk u so much!

Answer 14

Actually my internet is not stable

Answer 15

no problem :)
welcome ^_^