# Mathematics for Technology II (Math 2131)

Durham College, Mathematics
Free
• 36 lessons
• 0 quizzes
• 14 week duration
• ##### Solving Systems of Equations

This unit introduces how to systematically solve a system of equations, namely linear equations. Examples of non-linear systems, including systems of 3 unknowns will be of emphasis.

• ##### Graphs of Trigonometric Functions

The unit focuses primarily on how to graph periodic sinusoidal functions, and how to identify features of a waveform to produce an equation by inspection.

• ##### Polar Coordinate Functions

An introduction to the polar coordinate system.

• ##### Complex Numbers

This unit is an extension of what was introduced in Math 1131. To learn how to work with radicals, knowing your exponent laws in crucial. Hence, this unit begins with a thorough review.

• ##### Logarithmic Functions

This chapter introduces you to exponential functions, and how they can be solved using logarithms.

• ##### Trigonometric Identities and Equations
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• ##### Analytic Geometry
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## Mathematics for Technology II (Math 2131)

A radical is said to be in simplest form when:

1. The radicand has been reduced as much as possible.
2. There are no radicals in the denominator and no fractional radicands.
3. The index has been made as small as possible.

This section will cover part (1) only. For information on part (2) and part (3), follow the links.

In case you missed it, the anatomy of a radical looks like this:

Let’s begin with the simplest of radicals, the square root. To simplify these radicands, you breakdown the number or variables down into perfect squares if they’re not already perfect squares. Here are some examples:

Example 1 (very easy):

$\sqrt{49}\phantom{\rule{0ex}{0ex}}$

Nothing special needs to be done here: 49 is a perfect square, hence the square root is:

$\overline{)±7}\phantom{\rule{0ex}{0ex}}$

Example 2 (easy):

$\sqrt{49{x}^{2}}\phantom{\rule{0ex}{0ex}}$

This time the radicand consists of two perfect squares: 49 and x² as factors. You may distribute the radical to each factor if you like (√49 and √x²). 49 becomes ±7 and x² becomes x. Therefore, your answer is:

$\overline{)±7x}\phantom{\rule{0ex}{0ex}}$

Example 3 (medium):

$\sqrt{50{x}^{3}}\phantom{\rule{0ex}{0ex}}$

The radicand here consists of the factors: 50 and x³. Because these factors are not perfect squares, you can break down each individually into factors that are perfect squares. For example, 50 can be made into 25 and 2, while x³ made into x² and x.

• Always look for factors that are perfect n roots, where n represents the index of the radical:
$\sqrt{25·2·{x}^{2}·x}\phantom{\rule{0ex}{0ex}}\sqrt{25}\sqrt{2}\sqrt{{x}^{2}}\sqrt{x}\phantom{\rule{0ex}{0ex}}$

Notice now that √25 is ±5 and √x² is ±x. The rest √2 and √x can be placed underneath one radical:

$\overline{)±5x\sqrt{2x}}\phantom{\rule{0ex}{0ex}}$

Example 4 (hard):

$\sqrt{24{y}^{5}}\phantom{\rule{0ex}{0ex}}$

Once again, both of these factors are not perfect squares. Ask yourself, what can we break the 24 down into? What about the y⁵? The only combination of factors that multiplies to 24  – where one of the numbers is a perfect square is 6 and 4. For y⁵, that can be changed into y⁴ and y (explained below).

$\sqrt{6·4·{y}^{4}·y}\phantom{\rule{0ex}{0ex}}\sqrt{4}\sqrt{{y}^{4}}\sqrt{6}\sqrt{y}\phantom{\rule{0ex}{0ex}}\overline{)±2{y}^{2}\sqrt{6y}}\phantom{\rule{0ex}{0ex}}$
• The factor y⁴ is a perfect square because y² × y² makes y⁴. Another way to think of it is take y⁴ and raise it to the fractional exponent ½. The exponent rule (power raised to a power) would mean 4 × ½ = 2, as in y².

Example 5 (challenging):

$\sqrt{4{x}^{2}y+12{x}^{4}z}\phantom{\rule{0ex}{0ex}}$

Given that the radicand is a binomial, you cannot distribute the radical to both terms. In addition, this is the first time the radicand is anything other than a monomial. In situations like these, you want to make radicand into a monomial by factoring. Remember that factoring reduces the amount of terms in an expression while expanding does the opposite. Notice how 4 and x² are both common factors, hence they can be factored out.

$\sqrt{4{x}^{2}\left(y+3{x}^{2}z\right)}\phantom{\rule{0ex}{0ex}}\sqrt{4{x}^{2}}\sqrt{y+3{x}^{2}z}\phantom{\rule{0ex}{0ex}}\overline{)2x·\sqrt{y+3{x}^{2}z}}$

Now let’s concentrate on radicals whose index is greater than 2, such as cube roots, fourth, fifth roots, etc. If your radical is a cube root, for example, try to remove perfect cubes from the radicand. Similarly, if the radical is a fourth root, you look for perfect fourth roots, and so on.

Example 1:

$\sqrt[3]{24{y}^{5}}\phantom{\rule{0ex}{0ex}}$

Out of the box, 24 nor y⁵ are perfect cubes. Breaking 24 down into 8 and 3, however, makes one of the factors – 8 – a perfect cube (∛8 = 2). Similarly, y⁵ → y³ · y², where y³ becomes y after cube rooting.

$\sqrt[3]{8·3·{y}^{3}·{y}^{2}}\phantom{\rule{0ex}{0ex}}\sqrt[3]{8}\sqrt[3]{{y}^{3}}\sqrt[3]{3}\sqrt[3]{{y}^{2}}\phantom{\rule{0ex}{0ex}}\overline{)2y·\sqrt[3]{3{y}^{2}}}\phantom{\rule{0ex}{0ex}}$

Example 2:

$3\sqrt[5]{32x{y}^{11}}\phantom{\rule{0ex}{0ex}}$

See any perfect fifth roots? The fifth root of 32 is 2 – easy! On the contrary, x is hopeless, while y¹¹ can be broken down in y⁵, y⁵, and y (or y¹⁰ and y). y¹⁰ is a perfect fifth root than reduces to y². Therefore:

$3\sqrt[5]{{32}x{{y}}^{10}y}\phantom{\rule{0ex}{0ex}}3·{2}·{{y}}^{2}·\sqrt[5]{yx}\phantom{\rule{0ex}{0ex}}\overline{)6{y}^{2}·\sqrt[5]{yx}}\phantom{\rule{0ex}{0ex}}$