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Math Help - find the 2nd root

  1. #1
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    Smile find the 2nd root

    X^2-11x+k=0 one of the roots=5 find the other root.
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  2. #2
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    Hello,
    Quote Originally Posted by blame_canada100 View Post
    X^2-11x+k=0 one of the roots=5 find the other root.
    If a and b are the roots of a polynomial x+mx+n=0, then x+mx+n=(x-a)(x-b)
    If you develop the RHS, you'll get x-(a+b)x+ab.
    So you can see that m=-(a+b)

    Applying it here, you'll get -11=-(5+other root)
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  3. #3
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    Hi,

    One place to start is to use the basic quadratic formula:

    x=(-b+/-(b^2-4ac)^0.5)/2a.

    With a=1, b=-10, and c=k, and letting x=5 (as one of the roots) gives you k=25.

    Substituting this back into the question gives you x^2-10x+25=0. This is easily factorised to give x=5, twice.

    So the answer to the question is that the other solution is also 5.
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  4. #4
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    you can use discriminant X(squared)-11x+k=aX(squared)+bx+c

    a=1, b=-11, c=k, and let's say the discriminant D

    D=b(squared)-4ac
    =121-4*1*k
    =121-4k

    solutions are s1 and s2

    S1=(-b+(square root)D)/2a

    =(-11+(square root)121-4k)/2

    and S1=5...so
    (-11+(square root)121-4k)/2=5
    -11+(square root)121-4k=10

    (square root)121-4k=21
    now you put everywhere by squared and you get
    121-4k=441
    -4k=320
    k=-80

    so now let's look for S2
    D=(square root)121-4(-80)
    D=(square root)121+320
    D=(square root)441
    D=21

    S2=(-b-D)/2a
    S2=(-11-21)/2
    S2=-16
    the second root is -16
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