Question Solve for x; \\frac{1}{x+1}) + \\frac{2}{x+2}) = \\frac{4}{x+4}), x ≠-1,-2,-4.
Video Explanation
Explanatory Answer
Take (x + 1) (x + 2) as the common denominator of the left- hand side of the equation
\\frac{(x+2)+2(x+1)}{(x+1)(x+2)}) = \\frac{4}{x+4})
\\frac{x+2+2x+2}{(x+1)(x+2)}) = \\frac{4}{x+4})
\\frac{3x+4}{(x+1)(x+2)}) = \\frac{4}{x+4})
Cross multiply
(3x+4) (x+4) =4(x+1) (x+2)
3x2 + 12x + 4x + 16 = 4x2 + 8x + 4x + 8
3x2 + 16x + 16 = 4x2 + 12x + 8
x2 - 4x - 8 = 0
The quadratic equation cannot be factorized. So, let us use the formula to compute the roots of the equation.
Using the formula \\frac{-b± {\sqrt{b^2-4ac}}}{2a}) to find the root we get
\\frac{4 ± {\sqrt{(16-4(1)(-8)}}}{2})
= \\frac{4 ± {\sqrt{16 + 32}}}{2})
= \\frac{4 ± 4{\sqrt3}}{2}) = 2 ± 2\\sqrt3)