Question

$\frac{1}{x + 5 } + \frac{3}{12x + 4} = \frac{1}{x + 2}$
please solve this ​

Answer 1

$\huge\sf{Answer:-}$

2 and 3 are the roots of equation.

Step-by-step explanation:

Refer the above attachment.

Hope it helps!

Mark As Brainliest!

Answer 2

Answer :-

Roots of the equation are 2 and 3.

Explanation :-

$\dfrac{1}{x + 5} + \dfrac{3}{12x + 4} = \dfrac{1}{x + 2} \\ \\ \\ \implies \dfrac{1}{x + 2} - \dfrac{1}{x + 5} = \dfrac{3}{12x + 4} \\ \\ \\ \tt{taking \ lcm} \\ \\ \implies \dfrac{x + 5 - (x + 2)}{(x + 2)(x + 5)} = \dfrac{3}{12x + 4} \\ \\ \\ \implies \dfrac{x + 5 - x - 2}{x(x + 5) + 2(x + 5)} = \dfrac{3}{12x + 4} \\ \\ \\ \implies \dfrac{3}{ {x}^{2} + 5x + 2x + 10 } = \dfrac{3}{12x + 4} \\ \\ \\ \implies \dfrac{1}{ {x}^{2} + 7x + 10} = \dfrac{1}{12x + 4} \\ \\ \\ \implies {x}^{2} + 7x + 10 = 12x + 4 \\ \\ \\ \implies {x}^{2} + 7x + 10 - 12x - 4 = 0 \\ \\ \\ \implies {x}^{2} - 5x + 6 = 0 \\ \\ \\ \implies {x}^{2} - 2x - 3 x + 6 = 0 \\ \\ \\ \implies x(x - 2) - 3(x - 2) = 0 \\ \\ \\ \implies (x - 2)(x - 3) = 0 \\ \\ \\ \implies x - 2 = 0 \ \ or \ \ x - 3 = 0 \\ \\ \\ \implies x = 2 \ \ or \ \ x = 3$

∴ Roots of the equation are 2 and 3.