Question

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

x = negative 4 over 3 and x = −5
x = negative 4 over 3 and x = 5
x = 4 over 3 and x = −5
x = 4 over 3 and x = 5

Answer 1

x = negative 4 over 3 and x = 5

which is also B

Answer 2

we have

`(2)/(x-2)+ (7)/(x^(2)-4)= (5)/(x)`

Remember that

`x^(2)-4=(x+2)(x-2)`

Multiply by
`[x(x^(2) -4)]` both sides

`[x(x^(2) -4)](2)/(x-2)+[x(x^(2) -4)](7)/(x^(2)-4)= [x(x^(2) -4)](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(+/-)√(121+240)} {6}`

`x=\frac{11(+/-)√(361)} {6}`

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

`x=(11+19)/(6)=5`

`x=(11-19)/(6)=-(4)/(3)`

therefore

the answer is the option

x = negative 4 over 3 and x = 5