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purplemexican

  • 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.

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  1. DanJS
    • one year ago
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    dividing by zero breaks maths

  2. DanJS
    • one year ago
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    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. Purplemexican
    • one year ago
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    i tried that i still didnt understand

  4. DanJS
    • one year ago
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    If you have an equation like \[\frac{ x^2 }{ (x+2 )} = \frac{ 4 }{ (x+2) }\]

  5. DanJS
    • one year ago
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    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
    • one year ago
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    multiply both sides by x+2 gives \[x^2 = 4\] \[x = \sqrt{4} = \pm2\]

  7. DanJS
    • one year ago
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    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. Purplemexican
    • one year ago
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    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
    • one year ago
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    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. Purplemexican
    • one year ago
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    oh

  11. DanJS
    • one year ago
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    For the above example, the x=-2, causes (x+2) to be zero

  12. DanJS
    • one year ago
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    we can do another example if you want, there is one from that same site i linked

  13. Purplemexican
    • one year ago
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    sure

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