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

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Answer 1 · 2016-05-23T15:46:35

$x= 5" and " -4/3$

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

So $x^2-4->x^2-2^2 -> (x-2)(x+2)$

Write the given equation as:

$2/(x-2)+7/[(x-2)(x+2)]=5/x$

$=>[2(x+2)+7]/[(x-2)(x+2)]=5/x$

$=>[2x+4+7]/[(x-2)(x+2)]=5/x$

Cross multiplying

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

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

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

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

not all the solutions are whole numbers

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

$x=(-b+-sqrt(b^2-4ac))/(2a)$

$=>x=(11+-sqrt(11^2-4(3)(-20)))/(2(3))$

$x=(11+-sqrt(121+240))/6$

$x= (11+-19)/6$

$x= 5" and " -4/3$

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