Question

Solve the following quadratic equation:a) 1/(a+b+x) = 1/a+1/b+1/x , (a+b ≠0)

Answer 1

Given :

1/(a+b+x) = 1/a+1/b+1/x  
1/(a+b+x) - 1/x  = 1/a+1/b

x - (a+b+x) /x (a+b+x) = a+b / ab
[ Taking LCM  x (a+b+x) in LHS & ab in RHS]

x - a- b- x /x (a+b+x) = a+b / ab
-(a+b) / x (a+b+x) =  a+b / ab
-(a+b) × ab = x (a+b+x) × (a+b)
-(a+b) × ab - x (a+b+x) × (a+b) = 0
-(a+b){ ab + x (a+b+x) } = 0

[ Taking -(a+b) Common]
ab  + x (a+b+x )= 0      [ a+b ≠0]
ab + ax + bx +x² = 0
x² + ax +bx +ab = 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
Hence, x = -a or  x= -b.

HOPE THIS WILL HELP YOU...

Answer 2

$\frac{1}{a + b + x} = \frac{1}{a} + \frac{1}{b} + \frac{1}{x} \\ \frac{1}{a + b + x} - \frac{1}{x} = \frac{1}{a} + \frac{1}{b} \\ \frac{x - a - b - x}{x(a + b + x)} = \frac{a + b}{ab} \\ \frac{ - (a + b)}{x(a + b + x)} = \frac{(a + b)}{ab} \\ - ab = x(a + b + x) \\ {x}^{2} + (a + b)x + ab = 0 \\ {x}^{2} +a x + bx + ab = 0 \\ x(x + a) + b(x + a) = 0 \\ (x + a)(x + b) = 0 \\ x = - a \\ x = - b$