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Math Help - quadratic equation problem

  1. #1
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    quadratic equation problem

    x^2+p=4x+2 is three time the other roots.Find the value of p.


    please help..i dont know how to do..
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  2. #2
    Senior Member Dinkydoe's Avatar
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    Can you be a little more specific?

    Do you mean that if x=a is a root, then the other root is x=3a?

    Then find the corresponding p?
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  3. #3
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    Quote Originally Posted by Dinkydoe View Post
    Can you be a little more specific?

    Do you mean that if x=a is a root, then the other root is x=3a?

    Then find the corresponding p?

    yes.
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  4. #4
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    in that case, rewriting the equation as: x^2-4x+p-2=0 we have -4=4a and p-2=3a^2
    thus a=-1 and p=5
    the equation becomes x^2-4x+3=0 which has 1 and 3 as roots
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  5. #5
    Senior Member Dinkydoe's Avatar
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    Ok, write the equation as x^2-4x+(p-2)= 0

    Suppose that x^2-4x+(p-2) = (x-a)(x-3a)= x^2-4ax+3a^2

    Now we have a=1, 3a=3 and we can solve p-2=3a^2.
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  6. #6
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    Quote Originally Posted by mastermin346 View Post
    x^2+p=4x+2 One root is three times the other root. Find the value of p.


    please help..i dont know how to do..
    Hi mastermin346,

    Factorising is the fastest way.

    (x-root_1)(x-root_2)=0,\ x(x-root_2)-root_1(x-root_2)=0,

    x^2-(sum\ of\ roots)x+product\ of\ roots=0

    Here the sum of the roots=4.
    One is 3 times the other, so the sum of the roots is 4 times the smaller root.
    Smaller root = 4/4.

    If you had an expression that was difficult to factorise,
    you could proceed as follows.....

    The solutions (roots) of a quadratic equation ax^2+bx+c=0

    are x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

    One root is \frac{-b+\sqrt{b^2-4ac}}{2a}

    The other root (a quadratic has at most 2 roots) is \frac{-b-\sqrt{b^2-4ac}}{2a}

    First re-arrange your equation to ax^2+bx+c=0


    x^2+p=4x+2\ \Rightarrow\ (1)x^2+(-4)x+(p-2)=0

    a=1

    b=-4

    c=p-2

    If one root is 3 times the other, the bigger root must be the one with the +

    \frac{-b+\sqrt{b^2-4ac}}{2a}=3\left(\frac{-b-\sqrt{b^2-4ac}}{2a}\right)

    The denominator is common, so it is irrelevant.

    -b+\sqrt{b^2-4ac}=3\left(-b-\sqrt{b^2-4ac}\right)

    4+\sqrt{16-4(p-2)}=3\left(4-\sqrt{16-(p-2)}\right)

    \sqrt{16-4(p-2)}+3\sqrt{16-4(p-2)}=12-4=8

    4\sqrt{16-4(p-2)}=8

    This gives \sqrt{16-4(p-2)}=2

    16-4(p-2)=4

    12=4(p-2)

    3=p-2
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