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Math Help - quadradic problem

  1. #1
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    Exclamation quadradic problem

    The question is: determine the value(s) of m in each equation so that the two roots are equal. My teacher told me to use a discriminant. What's that?

    x^2-6x+m=o
    4x^2+mx+25

    Determine the value(s) of n in each equation such that one root is triple the other root.

    3x^2-4x+n=0
    4x^2+nx+27=0

    Help would be GREATLY appreaciated!!!

    Thank you!
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  2. #2
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    Quote Originally Posted by nathan02079 View Post
    The question is: determine the value(s) of m in each equation so that the two roots are equal. My teacher told me to use a discriminant. What's that?

    x^2-6x+m=o
    4x^2+mx+25

    Determine the value(s) of n in each equation such that one root is triple the other root.

    3x^2-4x+n=0
    4x^2+nx+27=0

    Help would be GREATLY appreaciated!!!

    Thank you!
    The quadratic equation

    ax^2+bx+c=0

    has the solutions:

    x = -\dfrac b{2a} \pm \dfrac1{2a} \sqrt{\underbrace{b^2-4ac}_{discriminant}}

    The discriminant is the term under the root-sign.

    According to your first question you get:

    x^2-6x+m=0~\implies~x=3\pm\sqrt{9-m}

    To get only one result the discriminant must be zero. Therefore m must be 9. Then both roots are 3.

    Do the second example in just the same way.

    =================================================
    Solve 3x^2-4x+n=0 for x:

    x=-\dfrac{-4}{2\cdot 3}\pm \dfrac1{6} \sqrt{16-4 \cdot 3 \cdot n}

    x_1=\dfrac23-\dfrac16\sqrt{16-12n}~\vee~x_2=\dfrac23+\dfrac16\sqrt{16-12n}

    According to the question you know that x_2 = 3\cdot x_1:

    \dfrac23+\dfrac16\sqrt{16-12n} = 3\cdot \left( \dfrac23-\dfrac16\sqrt{16-12n} \right)

    \dfrac23+\dfrac16\sqrt{16-12n} = 2-\dfrac12\sqrt{16-12n}

    \dfrac23\sqrt{16-12n} = \dfrac43

    \sqrt{16-12n} = 2

    16-12n = 4~\implies~\boxed{n=1}
    ==================================================
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