Question

1
$\div a + b + x = 1 \div a + 1 \div b + 1 \div x$
calculate using factorization method.​

Answer 1

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

Given equation is

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

can be rewritten as

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

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

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

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

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

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

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

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

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

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

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

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

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

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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

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

Given equation is

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

can be rewritten as

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

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

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

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

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

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

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

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

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

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

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

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

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

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

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