Answer:
Option A is correct.
i.e. x = 1, x = 0 is an extraneous solution.
Step-by-step explanation:
Given the expression
[tex]\frac{5}{x}=\frac{4x+1}{x^2}[/tex]
Solving the rational function
[tex]\frac{5}{x}=\frac{4x+1}{x^2}[/tex]
Apply fraction across multiply: if [tex]\frac{a}{b}=\frac{c}{d}\mathrm{\:then\:}a\cdot \:d=b\cdot \:c[/tex]
[tex]5x^2=x\left(4x+1\right)[/tex]
Subtract x(4x+1) from both sides
[tex]5x^2-x\left(4x+1\right)=x\left(4x+1\right)-x\left(4x+1\right)[/tex]
Simplify
[tex]5x^2-x\left(4x+1\right)=0[/tex]
5x² - 4x² - x = 0
x² - x = 0
Factor x² - x = x(x-1)
so
x(x-1) = 0
Using the zero factor principle
if ab=0, then a=0 or b=0 (or both a=0 and b=0)
[tex]x=0\quad \mathrm{or}\quad \:x-1=0[/tex]
Thus, the solution to the equation is:
[tex]x=0,\:x=1[/tex]
But, it is clear that if we substitute x = 0, the equation becomes undefined because we can not have the denominator to be 0.
In other words, the equation is undefined for x = 0
Thus, x = 0 is an extraneous solutions.
Therefore, option A is correct.
i.e. x = 1, x = 0 is an extraneous solution.
