Respuesta :
Extraneous solution:
An extraneous solution is a solution that arises from the solving process that is not really a solution at all
So, firstly we will solve for the equation
and then we verify each solutions by plugging them back
If denominator of rational equation becomes zero , then that solution must be extraneuous solution
For example:
[tex]\frac{1}{x+3} +\frac{2}{x} =-\frac{3}{x(x+3)}[/tex]
we can solve for x
Multiply both sides by x(x+3)
[tex]\frac{1}{x+3}x\left(x+3\right)+\frac{2}{x}x\left(x+3\right)=-\frac{3}{x\left(x+3\right)}x\left(x+3\right)[/tex]
now, we can simplify it
[tex]x+2\left(x+3\right)=-3[/tex]
[tex]x+2x+6=-3[/tex]
[tex]3x=-9[/tex]
now, we can solve for x
[tex]x=-3[/tex]
now, we can check whether x=-3 is extraneous solution
we will plug back x=-3 into original
[tex]\frac{1}{-3+3} +\frac{2}{-3} =-\frac{3}{-3(-3+3)}[/tex]
[tex]\frac{1}{0} +\frac{2}{-3} =-\frac{3}{-3(0)}[/tex]
we can see that denominator becomes 0
so, x=-3 can not be solution
so, x=-3 is extraneous solution...........Answer
Answer:
If a solution results in zero when substituted into the denominator of the equation, the solution is extraneous.
