Question

$\tt Solve :$
$\tt \dfrac{1}{a+b+x} = \dfrac{1}{a}+\dfrac{1}{b} = \dfrac{1}{x}$
$\tt where \: \:a+b≠0$
Please don't give unnecessary answers.​

Answer 1

Appropriate Question :-

Solve for x :-

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

$\large\underline{\sf{Solution-}}$

The given equation is

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

can be rewritten as

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

On taking LCM on both sides, we get

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

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

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

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

On dividing both sides by a + b, we get

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

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

$\rm \: {x}^{2} + ax + bx = - ab$

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

$\rm \: ( {x}^{2} + ax )+ (bx + ab) = 0$

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

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

$\rm \: x + a = 0 \: \: or \: \: x + b = 0$

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

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

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

Answer 2

$\bold { \underline{Given,}}$

$\color{red}\large \sf \frac{1}{a + \: b + \: x} \: \: = \: \: \frac{1}{a} + \: \frac{1}{b} \: + \frac{1}{x}$

$\large \color{blue}\sf \implies \: \frac{1}{a\:+\:+\:b\:+x}\:-\:1x\:=\: \frac{1}{a}\:+\:{1}{b}$

$\large \color{pink} \: \sf \frac{x\:-\:a\:+\:b\:+x}{a\:+\:b\:+\:x}\:=\: \frac{b\:+\:a}{a}\:\:\:\:\:(By \: taking \: LCM)$

$\large\color{aqua} \sf \frac{x\:-\:a\:-\:b\:-x}{(a\:+\:b\:+\:x)x}\:=\: \frac{b\:+\:a}{ab}$

$\large \color{lime} \sf\frac{-a\:-\:b}{(a\:+\:b\:+\:x)}\:=\: \frac{b\:+\:a}{ab}$

$\large \sf\color{red}\frac{-(a\:+\:b)}{a\:+\:b\:+\:x}\:=\: \frac{b\:+\:a}{ab}$

${ \large \sf \color{blue}\frac{-1}{(a\:+\:b\:+\:x)x}\:=\: \frac{1}{ab}} \: \: \: \: \small \sf(Dividing\:both\:sides\:by\:a\:+\:b)$

$\sf \: x(x + \: a \: + b ) \: = \: - \: ab \\ \\ \sf {x}^{2} + ax + bx \: = \: - \: ab \\ \\ \sf \: {x}^{2} + ax \: + bx \: + ab \: = \: 0$

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

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

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

$\sf \: x + a \: = \: 0 \: \: \: | \: \: \: x + b \: = \: 0 \\ \\ \sf \: x \: = \: - a \: \: \: | \: \: \: x \: = \: - \: b$