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Math Help - How to solve algebraic surds by factorising?

  1. #1
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    How to solve algebraic surds by factorising?

    This is the answer in the book but I don't know how they got from

    x^2-5root(2)X+12=0

    to

    (x-3root(2))(x-2root(2))

    Help please?
    Thank you!
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  2. #2
    Junior Member
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    well, you can use the quadratic formula to find the roots of the equation, which the roots are
    [-b+sqrt(b^2-4ac)]/2a and [-b-sqrt(b^2-4ac)]/2a

    where a=1, b=-5root(2), c=12

    after you get the roots, rewrite your equation in the form of
    (x-root1)(x-root2)=0
    Last edited by Jhevon; February 6th 2009 at 08:31 PM. Reason: wrote the name of the formula used
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  3. #3
    MHF Contributor

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    First note that (x+x_0)(x+x_1)= x^2+ (x_0+ x_1)x+ x_0x_1. That means that in order to factor a polynomial of the form x^2+ ax+ b you look for numbers x_0 and x_1 such that x_0+ x_1= a and x_0x_1= b. In this particular example a= -5\sqrt{2} and b= 12 so you need to solve x_0+ x_1= -5\sqrt{2} and x_0x_1= 12. Since that product does NOT have a \sqrt{2} in it, it is clear that both x_0 and x_1 must have a factor of \sqrt{2}.

    So, now, we are thinking x_0= m\sqrt{2} and x_1= n\sqrt{2} for some numbers m and n. x_0+ x_1= (m+n)\sqrt{2}= -5\sqrt{2} and x_0x_1= 2mn= 12 so we now have m+ n= 5 and mn= 6. -2 and -3 satisfy those equations so we can take x_0= -2\sqrt{2} and x_1= -3\sqrt{2} and have x^2- 5\sqrt{2}x+ 12= (x- 2\sqrt{2})(x- 3\sqrt{2})
    Last edited by Isomorphism; February 7th 2009 at 07:50 AM. Reason: Fixed the Latex Error
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