Mathematics

Question


[tex]3m ^{2} x ^{2} + 2mx - 5 = 0[/tex]

Answer?
With detailed steps please. Thank you! ​

1 Answer

  • We could factor the usual way but let's try Dr. Po Shen Loh's "new" method.

    [tex] 3m^2x^2 + 2mx - 5 = 0[/tex]

    First step is to rewrite as a monic,

    [tex]x^2 + \dfrac{2m}{3m^2} x - \dfrac{5}{3m^2} = 0[/tex]

    [tex]x^2 + \dfrac{2}{3m} x - \dfrac{5}{3m^2} = 0[/tex]

    Now we need two numbers which add to 2/3m and multiply to -5/3m².  If they add to 2/3m they average to 1/3m so they're 1/3m-u and 1/3m+u for some u.  The product of those two is -5/3m² so we write:

    [tex]\left( \dfrac{1}{3m}-u \right)\left( \dfrac{1}{3m} +u \right) = -\dfrac{5}{3m^2}[/tex]

    [tex]\dfrac{1}{9m^2}-u^2 = -\dfrac{5}{3m^2}[/tex]

    [tex]u^2 = \dfrac{1}{9m^2} +\dfrac{5(3)}{(3)3m^2} = \dfrac{16}{9m^2}[/tex]

    [tex]u = \pm \dfrac{4}{3m}[/tex]

    So our equation

    [tex]0=x^2 + \dfrac{2m}{3m^2} x - \dfrac{5}{3m^2}[/tex]

    factors as

    [tex]0= \left( x +\dfrac{1}{3m}-\dfrac{4}{3m}\right) \left( x + \dfrac{1}{3m} + \dfrac{4}{3m}\right)[/tex]

    [tex]0= \left(x - \dfrac{1}{m}\right) \left(x +\dfrac{5}{3m} \right)[/tex]

    so has roots

    [tex]x= \dfrac{1}{m} \textrm{ or } x = -\dfrac{5}{3m}[/tex]

    The more standard factorization with the same result is

    [tex]0=(mx-1)(3mx+5)[/tex]

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