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maths

posted by on .

Consider the quadratic equation
m^2 x^2 + 2(2m-5)x + 8 = 0

a) If 2 is a root of the equation, find the possible values of m (there are four possible values)

b) For each value of m, find the other root

I have actually calculated the answer m=1 or m=-3 for (a), but I was told that there are four possible values of m. I was stuck on this part.

  • maths - ,

    since 2 is a root,
    4m^2 + 2(2m-5)(2) + 8 = 0
    4m^2 + 8m - 20 + 8 = 0
    4m^2 + 8m - 12 =
    m^2 + 2m - 3 = 0
    (m+3)(m-1) = 0
    m = -3, or m = 1

    if m = -3, we have
    9x^2 -22x + 8 = 0
    (x-2)(9x - 4) = 0
    x = 2 (our given) or x = 4/9

    if m = 1 , we have
    x^2 -6x + 8 = 0
    (x-2)(x-4) = 0
    x=2 or x=4

    Like you, I only get 2 values for m
    and each value of m yields 2 answers. (one doubling up)

    I also approached the problem using the sum and product of roots property.
    let the roots be a and b, (we have a quadratic in x)
    but we know one of them let b = 2

    sum of roots = a+2
    product of roots = 2a

    sum of roots = -2(2m-5)/m^2
    a+2 = (-4m + 10)/m^2 - 2
    a = (-4m +10 - 2m^2)/m^2

    product of roots = 8/m^2
    2a = 8/m^2
    a = 4/m^2

    4/m^2 = (-4m + 10 - 2m^2)/m^2
    4 = -4m + 10 - 2m^2
    2m^2 + 4m - 6 = 0
    m^2 + 2m - 3 = 0
    (m+3)(m-1) = 0
    m = -3 or m = 1

    same as above, and only 2 values of m

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