Respuesta :
Answer: The correct option is
(B) [tex]x=-\dfrac{4}{3},~~5.[/tex]
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 :
[tex]\dfrac{2}{x-2}+\dfrac{7}{x^2-4}=\dfrac{5}{x}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)[/tex]
We will be using the following formula to solve equation (i) :
[tex]a^2-b^2=(a+b)(a-b).[/tex]
The solution of equation (i) is as follows :
[tex]\dfrac{2}{x-2}+\dfrac{7}{x^2-4}=\dfrac{5}{x}\\\\\\\Rightarrow \dfrac{2}{x-2}+\dfrac{7}{(x+2)(x-2)}=\dfrac{5}{x}\\\\\\\Rightarrow \dfrac{2(x+2)+7}{(x+2)(x-2)}=\dfrac{5}{x}\\\\\\\Rightarrow \dfrac{2x+11}{x^2-4}=\dfrac{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=-\dfrac{4}{3},~5.[/tex]
Thus, the required solution of the given equation is
[tex]x=-\dfrac{4}{3},~~5.[/tex]
Option (B) is CORRECT.
