Question

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

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Answer 1

Answer :

x = 2 ± 2√3

Step-by-step explanation :

    $\sf \dfrac{1}{x+1}+\dfrac{2}{x+2}=\dfrac{4}{x+4} \\\\\\ \dfrac{1(x+2)+2(x+1)}{(x+1)(x+2)}=\dfrac{4}{x+4} \\\\\\ \dfrac{x+2+2x+2}{x(x+2)+1(x+2)}=\dfrac{4}{x+4} \\\\\\ \dfrac{3x+4}{x^2+2x+x+2} = \dfrac{4}{x+4} \\\\\\ \dfrac{3x+4}{x^2+3x+1}=\dfrac{4}{x+4} \\\\\\ (3x+4)(x+4)=4(x^2+3x+2) \\\\ 3x(x+4)+4(x+4)=4x^2+12x+8 \\\\ 3x^2+12x+4x+16=4x^2+12x+8 \\\\ 3x^2+16x+16=4x^2+12x+8 \\\\ 4x^2+12x+8-3x^2-16x-16=0 \\\\ x^2-4x-8=0$

  It is of the form ax² + bx + c = 0

 a = 1 , b = -4 , c = -8

⇒  x =  $\sf \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$

Substituting the values of a, b and c

  $\sf x=\dfrac{-(-4) \pm \sqrt{(-4)^2-4(1)(-8)}}{2(1)} \\\\ x=\dfrac{4 \pm \sqrt{16+32}}{2} \\\\ x=\dfrac{4 \pm \sqrt{16(1+2)}}{2} \\\\ x=\dfrac{4 \pm \sqrt{16(3)}}{2} \\\\ x=\dfrac{4 \pm 4\sqrt{3}}{2} \\\\\ x=2 \pm 2\sqrt{3}$

Therefore, x = 2 + 2√3 , 2 - 2√3