A High School Math Problem Which Is a Lot Easier Than It Looks
This problem at first glance looks hard, but with the right method it is remarkably easy to solve
One of the most satisfying types of math problems to take on are those which look very hard on first sight, but turn out to be fairly easily solvable as long as you know the right method to apply.
Here is such a problem. Take a quick look and then see if you know how to solve it before I dive into the method I used:
Simplify the following:
Surds are expressions involving root symbols. Working with surds is often a core part of the high school mathematics curriculum in many countries. For many who will move onto careers where mathematical calculations are important, the ability to work with surds offers a valuable shortcut to solve problems where the computational accuracy is less important than the theoretical solution. Expressing calculations in terms of surds avoids the need to employ calculators and software in deriving a solution to a problem.
A surd is a little like an algebraic expression, in the sense that the root symbols are treated like variables in how they are manipulated. One of the most common manipulations that can be used with a simple surd of the form a+b√x is to multiply it by its conjugate a-b√x. Doing so removes the square root component of the surd to form product a² — b²x.
[Technically, positive numbers x always have two square root values, ±√x. However, in handling surds, we regard √x as the positive value +√x, and we denote its negation as -√x].
In general, wherever possible, it is beneficial to work with surds that do not involve roots in the denominators. If the denominator of an expression is a simple surd, we can rationalize if by multiplying both the numerator and the denominator by the conjugate of the denominator. So if our surd is 1/(a+b√x), and we rationalize by multiplying both top and bottom by a-b√x, we are left with the equivalent rationalized surd (a-b√x)/(a² — b²x), so that our surd now only has a root in the numerator.