Question

Find the value of x. Plz help. Wrong answers will be reported.​

Answer 1

${\huge{\underline{\boxed{\red{\mathcal{Answer}}}}}}$

Given:

• $\dfrac{1}{x + 1} + \dfrac{1}{x + 2} = \dfrac{4}{x + 4}$

To Find:

• Value of x

Solution:

$\implies \dfrac{1}{x + 1} + \dfrac{1}{x + 2} = \dfrac{4}{x + 4} \\\\\implies \dfrac{(x + 2)+(x + 1)}{(x + 1)(x+2)} = \dfrac{4}{x + 4} \\\\\implies \dfrac{2x + 3}{(x^2 + x + 2x + 2)} = \dfrac{4}{x + 4} \\\\\implies \dfrac{2x + 3}{(x^2 + 3x + 2)} = \dfrac{4}{x + 4} \\\\\implies (2x + 3)\times (x + 4) = 4\times(x^2 + 3x + 2) \\\\\implies 2x^2 + 8x + 3x + 12 = 4x^2 + 12x + 8 \\\\\implies 4x^2 - 2x^2 + 12x - 11x + 8 - 12 = 0 \\\\\implies 2x^2 + x - 4 = 0 \\\\\implies x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \;\;\;\;Where, a=2,\:b=1,\:c=-4 \\\\\implies x = \dfrac{-1 \pm \sqrt{1^2 - 4(2)(-4)}}{2(2)} \\\\\implies x = \dfrac{-1 \pm \sqrt{1 + 32}}{4} \\\\\bold{\implies x = \dfrac{-1 \pm \sqrt{33}}{4}}$

Hope it helps!!