Question

Solve the quadratic equation
(1) 1/a+b+x=1/a+1/b+1/c

Answer 1

1/(a+b+x)=1/a + 1/b + 1/x
1/(a+b+x)=(bx+ax+ab)/abx
abx=abx+a2x+a2b+b2x+abx+ab2+bx2+ax2+abx
ax2+bx2+a2x+abx+abx+b2x+a2b+ab2=0
x2(a+b)+ax(a+b)+bx(a+b)+ab(a+b)=0
(a+b)(x2+ax+bx+ab)=0
since, a+b!=0
so, x2+ax+bx+ab=0
x(x+a)+b(x+a)=0
(x+a)(x+b)=0
x=-a,-b

Answer 2

Answer:

Given :

$\displaystyle{\dfrac{1}{a+b+x} = \dfrac{1}{a} +\dfrac{1}{b} +\dfrac{1}{x} }$

Now take 1 / x in L.H.S. side :

$\displaystyle{\dfrac{1}{a+b+x} - \dfrac{1}{x}= \dfrac{1}{a} +\dfrac{1}{b} }$

Taking L.C.M and solving it further :

$\displaystyle{\dfrac{1}{a+b+x} - \dfrac{1}{x}= \dfrac{1}{a} +\dfrac{1}{b} }\\\\\\\displaystyle{\dfrac{x-(a+b+x)}{(a+b+x)(x)} = \dfrac{b+a}{ab} }\\\\\\\displaystyle{\dfrac{-a-b}{(a+b+x)(x)} = \dfrac{b+a}{ab} }\\\\\\\displaystyle{\dfrac{-(a+b)}{(a+b+x)(x)} = \dfrac{(a+b)}{ab} }$

$\displaystyle{\dfrac{-1}{(a+b+x)(x)} = \dfrac{1}{ab} }\\\\\\\displaystyle{\dfrac{-ab}{(a+b+x)(x)} = 1}\\\\\\\displaystyle{(a+b+x)(x)+ab=0}$

x² + a x + b x + a b = 0

x ( x + a ) + b ( x + a ) = 0

( x + a ) ( x + b ) = 0

x + a = 0  or x + b = 0

x = - a or x = - b .

Therefore , the value of x is - a or - b .