Question

$3m ^{2} x ^{2} + 2mx - 5 = 0$

Answer?
With detailed steps please. Thank you! ​

Answer 1

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

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