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.

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- purplemexican

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- DanJS

dividing by zero breaks maths

- 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

- purplemexican

i tried that i still didnt understand

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## More answers

- DanJS

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

- DanJS

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

- DanJS

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

- 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.

- purplemexican

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

- 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

- purplemexican

oh

- DanJS

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

- DanJS

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

- purplemexican

sure

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