Answer: The correct option is
(B)
![x=-(4)/(3),~~5.](https://img.qammunity.org/2020/formulas/mathematics/high-school/y694w0ho29yne04db8otx1bigtm969l8i3.png)
Step-by-step explanation: We are given to solve the following equation for x :
"2 over quantity x minus 2 plus 7 over quantity x squared minus 4 equals 5 over x."
The above equation can be written as :
![(2)/(x-2)+(7)/(x^2-4)=(5)/(x)~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)](https://img.qammunity.org/2020/formulas/mathematics/high-school/b40wih3x53g7uar34uhjq45r05ssbtqq59.png)
We will be using the following formula to solve equation (i) :
![a^2-b^2=(a+b)(a-b).](https://img.qammunity.org/2020/formulas/mathematics/high-school/lcsih7gboginoqi55jmropf5qhuupz1kc4.png)
The solution of equation (i) is as follows :
![(2)/(x-2)+(7)/(x^2-4)=(5)/(x)\\\\\\\Rightarrow (2)/(x-2)+(7)/((x+2)(x-2))=(5)/(x)\\\\\\\Rightarrow (2(x+2)+7)/((x+2)(x-2))=(5)/(x)\\\\\\\Rightarrow (2x+11)/(x^2-4)=(5)/(x)\\\\\Rightarrow 2x^2+11x=5x^2-20\\\\\Rightarrow 5x^2-2x^2-11x-20=0\\\\\Rightarrow 3x^2-11x-20=0\\\\\Rightarrow 3x^2-15x+4x-20=0\\\\\Rightarrow 3x(x-5)+4(x-5)=0\\\\\Rightarrow (3x+4)(x-5)=0\\\\\Rightarrow 3x+4=0,~~~x-5=0\\\\\Rightarrow x=-(4)/(3),~5.](https://img.qammunity.org/2020/formulas/mathematics/high-school/p97itsxq62ydl8hn27j0ywstkyrsdowgj5.png)
Thus, the required solution of the given equation is
![x=-(4)/(3),~~5.](https://img.qammunity.org/2020/formulas/mathematics/high-school/y694w0ho29yne04db8otx1bigtm969l8i3.png)
Option (B) is CORRECT.