How do you solve $\frac{2}{x - 2} + \frac{7}{x^{2} - 4} = \frac{5}{x}$ $2/(x-2) + 7/(x^2-4) = 5/x$ ?

Question

2534 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-05-23T15:46:35

$x = 5 and - \frac{4}{3}$ $x= 5" and " -4/3$

Known: $a^{2} - b^{2} = (a - b) (a + b)$ $a^2-b^2=(a-b)(a+b)$

So $x^{2} - 4 \to x^{2} - 2^{2} \to (x - 2) (x + 2)$ $x^2-4->x^2-2^2 -> (x-2)(x+2)$

Write the given equation as:

$\frac{2}{x - 2} + \frac{7}{(x - 2) (x + 2)} = \frac{5}{x}$ $2/(x-2)+7/[(x-2)(x+2)]=5/x$

$\Rightarrow \frac{2 (x + 2) + 7}{(x - 2) (x + 2)} = \frac{5}{x}$ $=>[2(x+2)+7]/[(x-2)(x+2)]=5/x$

$\Rightarrow \frac{2 x + 4 + 7}{(x - 2) (x + 2)} = \frac{5}{x}$ $=>[2x+4+7]/[(x-2)(x+2)]=5/x$

Cross multiplying

$x (2 x + 11) = 5 (x - 2) (x + 2)$ $x(2x+11)=5(x-2)(x+2)$

$x (2 x + 11) = 5 (x^{2} - 4)$ $x(2x+11)=5(x^2-4)$

$2 x^{2} + 11 x = 5 x^{2} - 20$ $2x^2+11x=5x^2-20$

$3 x^{2} - 11 x - 20 = 0$ $3x^2-11x-20=0$

not all the solutions are whole numbers

Using: $y = a x^{2} + b x + c$ $y=ax^2+bx+c$ where a=3; b= -11; c=-20

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x=(-b+-sqrt(b^2-4ac))/(2a)$

$\Rightarrow x = \frac{11 \pm \sqrt{11^{2} - 4 (3) (- 20)}}{2 (3)}$ $=>x=(11+-sqrt(11^2-4(3)(-20)))/(2(3))$

$x = \frac{11 \pm \sqrt{121 + 240}}{6}$ $x=(11+-sqrt(121+240))/6$

$x = \frac{11 \pm 19}{6}$ $x= (11+-19)/6$

$x = 5 and - \frac{4}{3}$ $x= 5" and " -4/3$

How do you solve 2/(x-2) + 7/(x^2-4) = 5/x2x−2+7x2−4=5x2/(x-2) + 7/(x^2-4) = 5/x?