## anonymous one year ago What causes a solution to a rational equation to be an extraneous solution? When there is more than one solution, one of the solutions is extraneous. If a solution results in zero when substituted into the denominator of the equation, the solution is extraneous. If a solution results in a negative number when substituted into the denominator of the equation, the solution is extraneous. When a solution is a fraction, the solution is extraneous.

1. DanJS

dividing by zero breaks maths

2. DanJS

here are a couple short vids that probably are decent, they usually cover the basics... https://www.khanacademy.org/math/algebra2/rational-expressions/solving-rational-equations/v/extraneous-solutions-to-rational-equations

3. anonymous

i tried that i still didnt understand

4. DanJS

If you have an equation like $\frac{ x^2 }{ (x+2 )} = \frac{ 4 }{ (x+2) }$

5. DanJS

if x = 2, the denominators become zero, and that is not allowed. So X can not be 2... now try to solve it...

6. DanJS

multiply both sides by x+2 gives $x^2 = 4$ $x = \sqrt{4} = \pm2$

7. DanJS

The solutions from solving say that X can be +2 or -2. however, When X=-2, the original equation is undefined (zero in denominator) so x=-2 is an "extraneous": solution.

8. anonymous

so youre saying If a solution results in a negative number when substituted into the denominator of the equation, the solution is extraneous

9. DanJS

no, the denominator can not be zero. If one of the solutions for X causes the denominator of the original equation to be zero, then it is extraneous solution

10. anonymous

oh

11. DanJS

For the above example, the x=-2, causes (x+2) to be zero

12. DanJS

we can do another example if you want, there is one from that same site i linked

13. anonymous

sure