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Math Help - Find the values of m?

  1. #1
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    Find the values of m?

    Find the values of m for which the difference between the roots of the equation 16x^2 = mx - 2501 is 5.
    I don't know how to start.
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by womatama View Post
    Find the values of m for which the difference between the roots of the equation 16x^2 = mx - 2501 is 5.
    I don't know how to start.
    Let a and b be roots to this equation.

    That means that 16a^2=ma-2501 and 16b^2=mb-2501\implies {\color{red}16b^2-mb+2501}=0.

    However, we impose case that a-b=5\implies a=5+b.

    Therefore,

    \begin{aligned}16(b+5)^2=m(b+5)-2501 & \implies 16(b^2+10b+25)=m(b+5)-2501\\ & \implies 16b^2+160b+400=mb+5m-2501\\ &\implies {\color{red}16b^2-mb+2501}=5m-160b-400\\ &\implies 5m=160b+400\\ &\implies m=32b+80\end{aligned}

    This tells us that the value of m depends on the value of one of the roots.

    Does this make sense?
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  3. #3
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    Quote Originally Posted by womatama View Post
    Find the values of m for which the difference between the roots of the equation 16x^2 = mx - 2501 is 5.
    I don't know how to start.
    1. The quadratic equation
    ax^2+bx+c=0 has the solutions
    x_1=-\frac b{2a}+\sqrt{\frac{b^2-4ac}{4a^2}}~\vee~x_2=-\frac b{2a}-\sqrt{\frac{b^2-4ac}{4a^2}}

    2. The difference between the 2 solutions is

    x_1-x_2=2\sqrt{\frac{b^2-4ac}{4a^2}}

    3. Your equation becomes

    16x^2-mx+2501=0

    where a = 16, b = -m, c = 2501

    Therefore you have to solve for m

    2\sqrt{\frac{m^2-4\cdot 16 \cdot 2501}{4 \cdot 256}} = 5

    I've got m=-408
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