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    Question
    HLPaper 2

    Consider the equation mx2−(m+3)x+2m+9=0mx^2 - (m + 3)x + 2m + 9 = 0mx2−(m+3)x+2m+9=0, where m∈Rm ∈ ℝm∈R.

    1.

    Write down an expression for the product of the roots, in terms of mmm.

    [1]
    Verified
    Solution

    product of roots =2m+9m=\frac{2m+9}{m}=m2m+9​ A1

    2.

    Hence or otherwise, determine the values of m such that the equation has one positive and one negative real root.

    [3]
    Verified
    Solution

    recognition that the product of the roots will be negative M1 2m+9m<0\frac{2m+9}{m}<0m2m+9​<0 A1 critical values m=0,−92m=0,-\frac{9}{2}m=0,−29​ seen A1 −92<m<0-\frac{9}{2}<m<0−29​<m<0 A1

    3.

    Find for what values there is one distinct real root.

    [3]
    Verified
    Solution

    Setting discriminant to 0

    • Δ=b2−4ac\Delta = b^2-4acΔ=b2−4ac
    • (m+3)2−4m(2m+9)=0(m+3)^2-4m(2m+9)=0(m+3)2−4m(2m+9)=0

    Solving for mmm

    • m2+6m+9−8m2−36m=−7m2−30m+9=0m^2+6m+9-8m^2-36m=-7m^2-30m+9=0m2+6m+9−8m2−36m=−7m2−30m+9=0
    • m=30±900+252−14m=\frac{30\pm\sqrt{900+252}}{-14}m=−1430±900+252​​
    • m≈−4.57m\approx-4.57m≈−4.57 or 0.280.280.28

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