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Quadratic identities

  1. Oct 3, 2009 #1
    1. The problem statement, all variables and given/known data

    By substituting y for 9x^2 solve 81x^4 - 63x^2 + 10 = 0

    2. Relevant equations

    3. The attempt at a solution

    My attempt at a solution is:

    y^2 - 7y + 10
    (y-2)(y-5)

    therefore y = 2 y = 5 Can someone double check this?

    Cheers
     
  2. jcsd
  3. Oct 3, 2009 #2
    That's correct. Now solve for x by setting 9x2 equal to each solution of y.
     
  4. Oct 3, 2009 #3
    Many thanks for the reply?

    Is this to check the solution?

    It just seemed to easy for some reason - maths is not my strong point, might I add.

    Cheers

     
  5. Oct 3, 2009 #4
    Oh I see - are you then saying

    9x^2 = 2
    9x = +- Sqroot 2
    x = + - Sqroot 2/9


    9x^2 = 5
    9x = +- Sqroot 5
    x= + - Sqroot 5/9


    Cheers
     
  6. Oct 3, 2009 #5

    Mentallic

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    Homework Helper

    Yes that is correct :smile:
    You can even check for yourself by substituing those x values back into the original quadratic equation. You should find they work.

    The only reason it asked you to substitute [tex]y=9x^2[/tex] into the equation [tex]81x^4 - 63x^2 + 10 = 0[/tex] is because it makes it more simple and easy to see how it should be solved.

    Rather than substituing, you could've always factorized it as so:
    [tex]81x^4-63x^2+10=(9x^2-2)(9x^2-5)=0[/tex]
     
  7. Oct 3, 2009 #6
    √(ab) = √(a)√(b) ≠ a√(b) which is what you did between the first and second lines above.

    You need to divide by 9 first, then take the square root of both sides.
    [tex]9x^2 = 2 \rightarrow x^2 = \frac{2}{9} \rightarrow x = \pm\sqrt{\frac{2}{3}} = \pm\frac{\sqrt{2}}{3}[/tex]

    Or take the square root of both sides completely, then solve for x
    [tex]9x^2 = 2 \rightarrow \sqrt{9x^2} = \sqrt{2} \rightarrow 3x = \pm\sqrt{2} \rightarrow x = \pm\frac{\sqrt{2}}{3}[/tex]
     
  8. Oct 3, 2009 #7

    Mentallic

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    Homework Helper

    Oh when I skimmed through it I read sqroot 2/9 as [tex]\sqrt{\frac{2}{9}}[/tex]. I didn't believe simplifying was top priority.
    Thanks for spotting that Bohrok.
     
  9. Oct 4, 2009 #8
    Hi there

    Many thanks for both your assistance - it is greatly appreciated.

    Cheers P
     
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