Is there an easy way to solve algebra problems?
2 Answers
There is not an easy way, there is THE Way to solve algebraic equations. You learn the rules, you learn how to execute the rules and then you solve problems using the correct rules the correct way.
It depends...
Explanation:
Sometimes there are easier and harder ways to solve problems. Sometimes there are no easy ways.
That having been said, unless you have seen a method being used, it is probably not easy to you. Once you have seen it, it may then be easy to you.
Here's an interesting example:
Consider the cubic equation:
#t^3-21t-90 = 0#
We can solve this using Cardano's method...
Let
Then we have:
#0 = (u+v)^3-21(u+v)-90#
#color(white)(0) = u^3+v^3+3(uv-7)(u+v)-90#
Add the constraint
#u^3+343/u^3-90 = 0#
Multiply through by
#(u^3)^2-90(u^3)+343 = 0#
By the quadratic formula:
#u^3 = (90+-sqrt((-90)^2-4(1)(343)))/(2*1)#
#color(white)(u^3) = 45+-sqrt(8100-1372)/2#
#color(white)(u^3) = 45+-29sqrt(2)#
Hence we can find real root:
#t_1 = root(3)(45+29sqrt(2))+root(3)(45-29sqrt(2))#
and complex roots:
#t_2 = omega root(3)(45+29sqrt(2))+omega^2 root(3)(45-29sqrt(2))#
#t_3 = omega^2 root(3)(45+29sqrt(2))+omega root(3)(45-29sqrt(2))#
where
OK, that may have seemed a little painful - perhaps there was an easier way:
Given:
#t^3-21t-90 = 0#
By the rational roots theorem, any rational roots of this cubic are expressible in the form
So the only possible rational roots are:
#+-1, +-2, +-3, +-5, +-6, +-9, +-10, +-15, +-18, +-30, +-45, +-90#
With a bit of trial and error we find:
#color(blue)(6)^3-21(color(blue)(6))-90 = 216-126-90 = 0#
So
#t^3-21t-90 = (t-6)(t^2+6t+15)#
Then the zeros of the remaining quadratic factor can be found by completing the square:
#0 = t^2+6t+15#
#color(white)(0) = t^3+6t+9+6#
#color(white)(0) = (t+3)^2-(sqrt(6)i)^2#
#color(white)(0) = ((t+3)-sqrt(6)i)((t+3)+sqrt(6)i)#
#color(white)(0) = (t+3-sqrt(6)i)(t+3+sqrt(6)i)#
So:
#t = -3+-sqrt(6)i#
No only did that seem easier, the resulting expressions for the solutions are much simpler.
