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purplemexican

  • one year ago

what is an extraneous solution

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  1. Nnesha
    • one year ago
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    do you have a equation or just need an example ?

  2. Purplemexican
    • one year ago
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    i have an equation

  3. madhu.mukherjee.946
    • one year ago
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    an extraneous solution represents a solution, such as that to an equation, that emerges from the process of solving the problem but is not a valid solution to the original problem.

  4. Nnesha
    • one year ago
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    post that you have to plug in the x value into the equation to check if is it extraneous or not if the value of x give you equal sides then yes if not then no that's not an extraneous solution

  5. Purplemexican
    • one year ago
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    (45-3x)^1/2=x-9 is the equation i have

  6. madhu.mukherjee.946
    • one year ago
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    Extraneous solutions can arise naturally in problems involving fractions with variables in the denominator. For example, consider this equation: \frac{1}{x - 2} = \frac{3}{x + 2} - \frac{6x}{(x - 2)(x + 2)}\,. To begin solving, we multiply each side of the equation by the least common denominator of all the fractions contained in the equation. In this case, the LCD is (x - 2)(x + 2). After performing these operations, the fractions are eliminated, and the equation becomes: x + 2 = 3(x - 2) - 6x\,. Solving this yields the single solution x = −2. However, when we substitute the solution back into the original equation, we obtain: \frac{1}{-2 - 2} = \frac{3}{-2 + 2} - \frac{6(-2)}{(-2 - 2)(-2 + 2)}\,. The equation then becomes: \frac{1}{-4} = \frac{3}{0} + \frac{12}{0}\,. This equation is not valid, since one cannot divide by zero.

  7. Nnesha
    • one year ago
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    can you solve for x ?

  8. mathstudent55
    • one year ago
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    An extraneous solution is a solution that appears in the solving of an equation that is not a solution to the original equation. Sometimes, you need to square both sides of an equation to solve the equation. The process of squaring both sides creates a new equation that is similar but not necessarily completely equivalent to the original equation. Sometimes there is a solution to the squared equation that is not a solution to the original equation. This is an extraneous solution, and it must be discarded.

  9. Purplemexican
    • one year ago
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    the is no solutions that i have found

  10. Nnesha
    • one year ago
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    you can convert 1/2 exponent to square root \[\sqrt{45-3x}=x-9\]

  11. Nnesha
    • one year ago
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    okay but you need x value x=?? no solution means the value of xx is an extraneous solution

  12. madhu.mukherjee.946
    • one year ago
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    hey its coming to x^2-15x+36=0

  13. madhu.mukherjee.946
    • one year ago
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    x=12 0r 3

  14. madhu.mukherjee.946
    • one year ago
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    if two sides of the equation can be squared then its coming

  15. Purplemexican
    • one year ago
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    my answer choices are x = –12 x = –3 x = 3 x = 12

  16. madhu.mukherjee.946
    • one year ago
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    its 12 and 3 .i can show you my workings if you really want

  17. Nnesha
    • one year ago
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    sigh. :=)

  18. Purplemexican
    • one year ago
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    i have to pick one of them i cant have both

  19. madhu.mukherjee.946
    • one year ago
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    dear its a quadratic equation so you'll always have two answers

  20. Nnesha
    • one year ago
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    plugin the x value into the equation which one give you the equal sides ?

  21. madhu.mukherjee.946
    • one year ago
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    take 12

  22. Nnesha
    • one year ago
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    you need an*extraneous solution* \[\huge\rm \sqrt{45-3\color{reD}{x}}=\color{reD}{x}-9\] first replace x by 12 do you get equal sides and then replace by 3 do you get equal sdes ?

  23. Nnesha
    • one year ago
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    sides*

  24. mathstudent55
    • one year ago
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    Square both sides of your equation to get rid of the 1/2 exponent. \(\large (45-3x)^{\frac{1}{2}}=x-9\) \(\large [(45-3x)^{\frac{1}{2}}]^2=(x-9)^2\) \(\large 45-3x=x^2 - 18x + 81\) \(\large x^2 - 15x + 36 = 0\) \(\large(x - 12)(x - 3) = 0\) \(\large x = 12\) or \(\large x = 3\) 12 and 3 are the solutions to the quadratic equation we got after squaring both sides. The squaring of both sides can introduce extraneous solution, so both solutions to the quadratic, 3 and 12, must be checked in the original equation. When you check x = 12 in the original equation, it does work, so the only solution to the original equation is x = 12. When you use x = 3 in the original equation, you see that the right side becomes -6. A square root cannot be a negative number, so x = 3 must be discarded since it is an extraneous solution.

  25. madhu.mukherjee.946
    • one year ago
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    i told you if you want a proper solution then take 12 otherwise take 3

  26. Purplemexican
    • one year ago
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  27. Purplemexican
    • one year ago
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    it cant be 12 it has to be 3 if n extraneous solution is a invalid answer

  28. mathstudent55
    • one year ago
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    The extraneous solution is 3. The real solution is 12.

  29. madhu.mukherjee.946
    • one year ago
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    okay then its 3

  30. madhu.mukherjee.946
    • one year ago
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    choose solutions as per your requirement

  31. Purplemexican
    • one year ago
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    thank you to all who helped i appreciate it.

  32. madhu.mukherjee.946
    • one year ago
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    your welcome

  33. madhu.mukherjee.946
    • one year ago
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    :)))))))

  34. Nnesha
    • one year ago
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    now substitute 3 \[\huge\rm \sqrt{45-3\color{reD}{(3)}}=\color{reD}{3}-9\] \[\huge\rm \sqrt{36} =-6\] square root of 36 =6 but \[6\cancel{=}-6 \] both sides are not equal so 3 is ane extraneous

  35. Nnesha
    • one year ago
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    \[\huge\rm \sqrt{45-3\color{reD}{(12)}}=\color{reD}{12}-9\] \[\huge\rm \sqrt{9} =3\] now take square root of 9 both sides are equal then 12 is not an extraneous solution

  36. madhu.mukherjee.946
    • one year ago
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    @Nnesha an extraneous solution is an invalid one so how can you think of 12 .coz 12 is a perfect solution .3 is an extraneous solution coz its giving you an invalid answer:)))

  37. Nnesha
    • one year ago
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    corrected.

  38. Nnesha
    • one year ago
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    that was a typo

  39. madhu.mukherjee.946
    • one year ago
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    okay:)))

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