Question

FIND THE ROOTS:
$\frac{1}{x + 1} + \frac{1}{x + 2} = \frac{4}{x + 4} \\ x \: notequals \: to - 1 \: \: - 2 \: \: \: - 4$
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Answer 1

Answer:

1/x+1 + 1/x+2

= 1.499694891

4/x+4 = 0.9999389686

-1-2-4 = -7

this is the x value = -7

Answer 2

Question:

FIND THE ROOTS

$\frac{1}{x + 1} + \frac{1}{x + 2} = \frac{4}{x + 4} </p><p>$

Answer:

$\frac{(x + 2) + (x + 1)}{(x + 1)(x + 2)} = \frac{4}{x + 4} \\ \\ \frac{2x + 3}{(x + 1)(x + 2)} = \frac{4}{x + 4} \\ \\ (2x + 3)(x + 4) = 4(x + 1)(x + 2) \\ \\ 2 {x}^{2} + 8x + 3x + 12 = 4( {x}^{2} + 2x + x + 2) \\ \\ 2 {x}^{2} + 8x + 3x + 12 = 4 {x}^{2} + 8x + 4x + 2 \\ \\ - 2 {x}^{2} - x + 10 = 0 \\ \\ 2 {x}^{2} + x + 10 = 0$

Use factorisation method to find the root of the equation,

$2 {x}^{2} + 5x - 4x + 10 = 0 \\ \\ x(2x + 5) - 2(2x - 5) = 0 \\ \\ (x - 2)(2x + 5) = 0$

Take

$x - 2 = 0 \: \: \: \: \: \: \: \: \: 2x + 5 = 0 \\ \\ x = 2 \: \: \: \: \: \: x = \frac{ - 5}{2}$

Hence the root of the equation is x= 2,-5/2