Question

Solve for x: 2 over quantity x minus 2 plus 7 over quantity x squared minus 4 equals 5 over x.

Answer 1

we have

$\frac{2}{x-2}+ \frac{7}{x^{2}-4}= \frac{5}{x}$

Remember that

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

Multiply by $[x(x^{2} -4)]$ both sides

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

Simplify

$2x(x+2)+7x=5(x^{2}-4) \\ \\2x^{2}+4x+7x=5 x^{2}-20 \\ \\3x^{2}-11x-20=0$

The formula to solve a quadratic equation of the form $ax^{2} +bx+c=0$ is equal to

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

in this problem we have

$3x^{2} -11x-20=0$  

so

$a=3\\b=-11\\c=-20$

substitute in the formula

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

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

$x=\frac{11(+/-)\sqrt{361}} {6}$

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

$x=\frac{11+19}{6}=5$

$x=\frac{11-19}{6}=-\frac{4}{3}$

therefore

the answer is the option

x = negative 4 over 3 and x = 5

Answer 2

 x = negative 4 over 3 and x = 5

which is also B