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When simplifying, you won't always have only numbers inside the radical; you'll also have to work with variables. Variables in a radical's argument are simplified in the same way as regular numbers. You factor things, and whatever you've got a pair of can be taken 'out front'.
Simplify
I already know that 16 is 42, so I know that I'll be taking a 4 out of the radical. Looking then at the variable portion, I see that I have two pairs of x's, so I can take out one x from each pair. Then:
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Purple Notes 4 2 Notes
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As you can see, simplifying radicals that contain variables works exactly the same way as simplifying radicals that contain only numbers. We factor, find things that are squares (or, which is the same thing, find factors that occur in pairs), and then we pull out one copy of whatever was squared (or of whatever we'd found a pair of).
Simplify
Looking at the numerical portion of the radicand, I see that the 12 is the product of 3 and 4, so I have a pair of 2's (so I can take a 2 out front) but a 3 left over (which will remain behind inside the radical).
Looking at the variable portion, I have two pairs of a's; I have three pairs of b's, with one b left over; and I have one pair of c's, with one c left over. So the root simplifies as:
You are used to putting the numbers first in an algebraic expression, followed by any variables. But for radical expressions, any variables outside the radical should go in front of the radical, as shown above. Always put everything you take out of the radical in front of that radical (if anything is left inside it).
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Simplify
Writing out the complete factorization would be a bore, so I'll just use what I know about powers. The 20 factors as 4 × 5, with the 4 being a perfect square. The r18 has nine pairs of r's; the s is unpaired; and the t21 has ten pairs of t's, with one t left over. Then:
Technical point: Your textbook may tell you to 'assume all variables are positive' when you simplify. Why? Because the square root of the square of a negative number is not the original number.
For instance, you could start with –2, square it to get +4, and then take the square root of +4 (which is defined to be the positive root) to get +2. You plugged in a negative and ended up with a positive.
We're applying a process that results in our getting the same numerical value, but it's always positive (or at least non-negative). Sound familiar? It should: it's how the absolute value works: |–2| = +2. Taking the square root of the square is in fact the technical definition of the absolute value.
But this technicality can cause difficulties if you're working with values of unknown sign; that is, with variables. The |–2| is +2, but what is the sign on | x |? You can't know, because you don't know the sign of x itself — unless they specify that you should 'assume all variables are positive', or at least non-negative (which means 'positive or zero').
Multiplying Square Roots
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The first thing you'll learn to do with square roots is 'simplify' terms that add or multiply roots.
Simplifying multiplied radicals is pretty simple, being barely different from the simplifications that we've already done. We use the fact that the product of two radicals is the same as the radical of the product, and vice versa.
Write as the product of two radicals:
Because 6 factors as 2 × 3, I can split this one radical into a product of two radicals by using the factorization. (Yes, I could also factorize as 1 × 6, but they're probably expecting the prime factorization.)
Yes, that manipulation was fairly simplistic and wasn't very useful, but it does show how we can manipulate radicals. And using this manipulation in working in the other direction can be quite helpful. For instance:
Simplify by writing with no more than one radical:
When multiplying radicals, as this exercise does, one does not generally put a 'times' symbol between the radicals. The multiplication is understood to be 'by juxtaposition', so nothing further is technically needed.
To do this simplification, I'll first multiply the two radicals together. This will give me 2 × 8 = 16 inside the radical, which I know is a perfect square.
By the way, I could have done the simplification of each radical first, then multiplied, and then does another simplification. The work would be a bit longer, but the result would be the same:
sqrt[2] × sqrt[8] = sqrt[2] × sqrt[4] sqrt[2]
= sqrt[2] × 2 sqrt[2]
= 2 × sqrt[2] sqrt[2]
= 2 × 2 = 4
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Simplify by writing with no more than one radical:
Neither of the radicals they've given me contains any squares, so I can't take anything out front — yet. What happens when I multiply these together?
Simplify by writing with no more than one radical:
As these radicals stand, nothing simplifies. However, once I multiply them together inside one radical, I'll get stuff that I can take out, because:
6 × 15 × 10 = 2 × 3 × 5 × 2 × 5
So I'll be able to take out a 2, a 3, and a 5:
The process works the same way when variables are included:
Simplify by writing with no more than one radical:
Purple Notes 4 2 Vocab
The 4 in the first radical is a square, so I'll be able to take its square root, 2, out front; I'll be stuck with the 5 inside the radical. By multiplying the variable parts of the two radicals together, I'll get x4, which is the square of x2, so I'll be able to take x2 out front, too.
You can use the Mathway widget below to practice simplifying products of radicals. Try the entered exercise, or type in your own exercise. Then click the button to compare your answer to Mathway's.
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URL: https://www.purplemath.com/modules/radicals2.htm
Purplemath
In mathematics, an 'identity' is an equation which is always true. These can be 'trivially' true, like 'x = x' or usefully true, such as the Pythagorean Theorem's 'a2 + b2 = c2' for right triangles. There are loads of trigonometric identities, but the following are the ones you're most likely to see and use.
Basic & Pythagorean, Angle-Sum & -Difference, Double-Angle, Half-Angle, Sum, Product
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Basic and Pythagorean Identities
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Notice how a 'co-(something)' trig ratio is always the reciprocal of some 'non-co' ratio. You can use this fact to help you keep straight that cosecant goes with sine and secant goes with cosine.
The following (particularly the first of the three below) are called 'Pythagorean' identities.
sin2(t) + cos2(t) = 1
tan2(t) + 1 = sec2(t)
1 + cot2(t) = csc2(t)
Note that the three identities above all involve squaring and the number 1. You can see the Pythagorean-Thereom relationship clearly if you consider the unit circle, where the angle is t, the 'opposite' side is sin(t) = y, the 'adjacent' side is cos(t) = x, and the hypotenuse is 1.
We have additional identities related to the functional status of the trig ratios:
sin(–t) = –sin(t)
cos(–t) = cos(t)
tan(–t) = –tan(t)
Notice in particular that sine and tangent are odd functions, being symmetric about the origin, while cosine is an even function, being symmetric about the y-axis. The fact that you can take the argument's 'minus' sign outside (for sine and tangent) or eliminate it entirely (for cosine) can be helpful when working with complicated expressions.
Angle-Sum and -Difference Identities
sin(α + β) = sin(α) cos(β) + cos(α) sin(β)
sin(α – β) = sin(α) cos(β) – cos(α) sin(β)
cos(α + β) = cos(α) cos(β) – sin(α) sin(β)
cos(α – β) = cos(α) cos(β) + sin(α) sin(β)
By the way, in the above identities, the angles are denoted by Greek letters. The a-type letter, 'α', is called 'alpha', which is pronounced 'AL-fuh'. The b-type letter, 'β', is called 'beta', which is pronounced 'BAY-tuh'.
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Double-Angle Identities
sin(2x) = 2 sin(x) cos(x)
cos(2x) = cos2(x) – sin2(x) = 1 – 2 sin2(x) = 2 cos2(x) – 1
Half-Angle Identities
The above identities can be re-stated by squaring each side and doubling all of the angle measures. The results are as follows:
sin2(x) = ½[1 – cos(2x)]
cos2(x) = ½[1 + cos(2x)]
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Sum Identities
Product Identities
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You will be using all of these identities, or nearly so, for proving other trig identities and for solving trig equations. However, if you're going on to study calculus, pay particular attention to the restated sine and cosine half-angle identities, because you'll be using them a lot in integral calculus.
URL: https://www.purplemath.com/modules/idents.htm
