Question

solve x. 1/a+b+x=1/a+1/b+1/x​

Answer 1

Answer:

$\boxed{\sf \: \: x \: = \: - \: a \: , \: - \: b \: \: }$

Step-by-step explanation:

Given equation is

$\sf \: \dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{x} = \dfrac{1}{a + b + x}$

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

$\sf \: \dfrac{b + a}{ab} = \dfrac{x - (a + b + x)}{(a + b + x)x}$

$\sf \: \dfrac{b + a}{ab} = \dfrac{x - a - b - x}{(a + b + x)x}$

$\sf \: \dfrac{b + a}{ab} = \dfrac{ - a - b }{(a + b + x)x}$

$\sf \: \dfrac{b + a}{ab} = \dfrac{ - (a + b)}{(a + b + x)x}$

On cancel out (a + b) from both sides, we get

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

$\sf \: x(a + b + x) = - ab$

$\sf \: xa + xb + {x}^{2} = - ab$

$\sf \: {x}^{2} + ax + bx + ab = 0$

$\sf \: x(x + a) + b(x + a) = 0$

$\sf \: (x + a) \: (x + b) = 0$

$\implies\sf \: x \: = \: - \: a \: \: or \: \: x \: = \: - \: b$

Hence,

$\implies\sf \: x \: = \: - \: a \: , \: - \: b$

$\rule{190pt}{2pt}$

Additional Information

Nature of roots :

Let us consider a quadratic equation ax² + bx + c = 0, then nature of roots of quadratic equation depends upon Discriminant (D) of the quadratic equation.

If Discriminant, D > 0, then roots of the equation are real and unequal.

If Discriminant, D = 0, then roots of the equation are real and equal.

If Discriminant, D < 0, then roots of the equation are unreal or complex or imaginary.

Where,

Discriminant, D = b² - 4ac