[tex]3m ^{2} x ^{2} + 2mx  5 = 0[/tex] Answer? With detailed steps please. Thank you!
Question
[tex]3m ^{2} x ^{2} + 2mx  5 = 0[/tex]
Answer?
With detailed steps please. Thank you!
1 Answer

1. User Answers DeanR
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 1/3mu 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=(mx1)(3mx+5)[/tex]