Question

$\frac{1}{x + 1} + \frac{2}{x + 2} = \frac{4}{x + 4}$
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Answer 1

Answer:

This equation firstly will be reduced to a simple quadratic equation

as:-)

$\large \frac{1}{x + 1} + \frac{2}{x + 2} = \frac{4}{x + 4} \\ \\ \large \frac{x + 2 + 2x + 2}{(x + 1)(x + 2)} = \frac{4}{x + 4} \\ \\ \large\frac{3x + 4}{ {x}^{2} + 3x + 2} = \frac{4}{x + 4}$

Now,

cross multiplying we get,

$3 {x}^{2} + \cancel{12x} + 4x + 16 = 4 {x}^{2} + \cancel {12x} + 8 \\ \\ {x }^{2} - 4x - 8 = 0$

Now we can solve it as:-)

${x}^{2} - 4x - 8 = 0 \\ \\ x = \large\frac{ 4 \pm \sqrt{16 + 32} }{2} \\ \\ = \large \frac{4 \pm \sqrt{48} }{2} \\ \\ = \large \frac{4 \pm4 \sqrt{3} }{2} \\ \\ = \large 2(1 \pm \sqrt{3} )$

Which is reqrd solution