# Is there an easy way to solve algebra problems?

Mar 25, 2017

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.

Mar 25, 2017

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:

${t}^{3} - 21 t - 90 = 0$

We can solve this using Cardano's method...

Let $t = u + v$

Then we have:

$0 = {\left(u + v\right)}^{3} - 21 \left(u + v\right) - 90$

$\textcolor{w h i t e}{0} = {u}^{3} + {v}^{3} + 3 \left(u v - 7\right) \left(u + v\right) - 90$

Add the constraint $v = \frac{7}{u}$ so the $\left(u + v\right)$ term goes away and we get:

${u}^{3} + \frac{343}{u} ^ 3 - 90 = 0$

Multiply through by ${u}^{3}$ to get:

${\left({u}^{3}\right)}^{2} - 90 \left({u}^{3}\right) + 343 = 0$

${u}^{3} = \frac{90 \pm \sqrt{{\left(- 90\right)}^{2} - 4 \left(1\right) \left(343\right)}}{2 \cdot 1}$

$\textcolor{w h i t e}{{u}^{3}} = 45 \pm \frac{\sqrt{8100 - 1372}}{2}$

$\textcolor{w h i t e}{{u}^{3}} = 45 \pm 29 \sqrt{2}$

Hence we can find real root:

${t}_{1} = \sqrt[3]{45 + 29 \sqrt{2}} + \sqrt[3]{45 - 29 \sqrt{2}}$

and complex roots:

${t}_{2} = \omega \sqrt[3]{45 + 29 \sqrt{2}} + {\omega}^{2} \sqrt[3]{45 - 29 \sqrt{2}}$

${t}_{3} = {\omega}^{2} \sqrt[3]{45 + 29 \sqrt{2}} + \omega \sqrt[3]{45 - 29 \sqrt{2}}$

where $\omega = - \frac{1}{2} + \frac{\sqrt{3}}{2} i$

OK, that may have seemed a little painful - perhaps there was an easier way:

Given:

${t}^{3} - 21 t - 90 = 0$

By the rational roots theorem, any rational roots of this cubic are expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $- 90$ and $q$ a divisor of the coefficient $1$ of the leading term.

So the only possible rational roots are:

$\pm 1 , \pm 2 , \pm 3 , \pm 5 , \pm 6 , \pm 9 , \pm 10 , \pm 15 , \pm 18 , \pm 30 , \pm 45 , \pm 90$

With a bit of trial and error we find:

${\textcolor{b l u e}{6}}^{3} - 21 \left(\textcolor{b l u e}{6}\right) - 90 = 216 - 126 - 90 = 0$

So $t = 6$ is a root and $\left(t - 6\right)$ a factor:

${t}^{3} - 21 t - 90 = \left(t - 6\right) \left({t}^{2} + 6 t + 15\right)$

Then the zeros of the remaining quadratic factor can be found by completing the square:

$0 = {t}^{2} + 6 t + 15$

$\textcolor{w h i t e}{0} = {t}^{3} + 6 t + 9 + 6$

$\textcolor{w h i t e}{0} = {\left(t + 3\right)}^{2} - {\left(\sqrt{6} i\right)}^{2}$

$\textcolor{w h i t e}{0} = \left(\left(t + 3\right) - \sqrt{6} i\right) \left(\left(t + 3\right) + \sqrt{6} i\right)$

$\textcolor{w h i t e}{0} = \left(t + 3 - \sqrt{6} i\right) \left(t + 3 + \sqrt{6} i\right)$

So:

$t = - 3 \pm \sqrt{6} i$

No only did that seem easier, the resulting expressions for the solutions are much simpler.