Answer:
The equation that would be used for the zero product property on will be (x-5)(x-10)=0
Option B is correct.
Step-by-step explanation:
We need to solve the equation [tex]\frac{1}{(x-4)}+1=\frac{14}{x+2}[/tex]
Solving the equation:
[tex]\frac{1}{(x-4)}+1=\frac{14}{x+2}\\\frac{1+(x-4)}{x-4}= \frac{14}{x+2}\\\frac{1+x-4}{x-4}= \frac{14}{x+2}\\\frac{x-3}{x-4}= \frac{14}{x+2}\\Cross\: Multiply\\(x-3)(x+2)=14(x-4)\\x(x+2)-3(x+2)=14x-56\\x^2+2x-3x-6=14x-56\\x^2-x-6-14x+56=0\\x^2-x-14x-6+50=0\\x^2-15x+50=0[/tex]
Now, we would factorise to find value of x
[tex]x^2-15x+50=0\\x^2-5x-10x+50=0\\x(x-5)-10(x-5)=0\\(x-5)(x-10)=0[/tex]
So, the equation that would be used for the zero product property on will be (x-5)(x-10)=0
Option B is correct.
