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Makovka662 [10]
3 years ago
5

^{2} x ^{2} + 2mx - 5 = 0" alt="3m ^{2} x ^{2} + 2mx - 5 = 0" align="absmiddle" class="latex-formula">

Answer?
With detailed steps please. Thank you! ​
Mathematics
1 answer:
crimeas [40]3 years ago
8 0

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

3m^2x^2 + 2mx - 5 = 0

First step is to rewrite as a monic,

x^2 + \dfrac{2m}{3m^2} x - \dfrac{5}{3m^2} = 0

x^2 + \dfrac{2}{3m} x - \dfrac{5}{3m^2} = 0

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:

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

\dfrac{1}{9m^2}-u^2 = -\dfrac{5}{3m^2}

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

u = \pm \dfrac{4}{3m}

So our equation

0=x^2 + \dfrac{2m}{3m^2} x - \dfrac{5}{3m^2}

factors as

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

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

so has roots

x= \dfrac{1}{m} \textrm{ or } x = -\dfrac{5}{3m}

The more standard factorization with the same result is

0=(mx-1)(3mx+5)

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