Question

$\huge\bf {Todays\: Question:- }$

Topic :- Quadratic fun !

A root of the equation $\bf \dfrac{a+c}{x+a} +\dfrac{b+c}{x+b } = \dfrac{2(a+b+c)}{x+a+b}$

Options :-
a) a
b) b
c) c
d) a, b, c

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Answer 1

$\implies$$\bf \dfrac{a+c}{x+a} +\dfrac{b+c}{x+b } = \dfrac{2(a+b+c)}{x+a+b}$

${\small{\sf{=>(x+b)(a+c)+(x+a)(b+c)}}}$$\small\sf{=\dfrac{2(a + b + c)(x + a)(x + b)}{(x + a + b)}}$

$\small\sf{=>{x}^{2} (a+b+2c)+x\{2ab+(a+b)\}c= {2x}^{2} (a+b+c)}$$\small\sf{+2(a+b)x(a+b+c)2x ^{2} (a+b)+}$$\small\sf{2x\{(a−b)(a+b+c)\}+(a+b−c)(a+b)=0}$

$\therefore\boxed{\bf{c) \:c\: is\: the\: right \:option}}$