This section covers:

**Introducing Exponents and Radicals (Roots) with Variables****Properties of Exponents and Radicals**,**Putting Exponents and Radicals in the Calculator****Rationalizing Radicals****Simplifying Exponential Expressions****Solving Exponential and Radical Equations****Solving Simple Radical Inequalities****More Practice**

We briefly talked about exponents in the **Powers, Exponents, Radicals (Roots) and Scientific Notation** section, but we need to go a little bit further in depth and talk about how to do algebra with them. Note that we’ll see more radicals in the **Solving Radical Equations and Inequalities **section, and we’ll talk about **Factoring with Exponents**, and** Exponential Functions** in the **Exponential Functions** section. Remember that **exponents**, or “raising” a number to a power, are just the number of times that the number (called the **base**) is multiplied by itself. **Radicals** (which comes from the word “root” and means the same thing) means undoing the exponents, or finding out what numbers **multiplied by themselves** comes up with the number. So we remember that* *\(\sqrt{25}=5\)*, *since \(5\times 5=25\)*. *Note that we have to remember that when taking the square root (or any even root), we always take the **positive value** (just memorize this).

# Introducing Exponents and Radicals (Roots) with Variables

But now that we’ve learned some algebra, we can do exponential problems with variables in them!* *So we have \(\sqrt{{{x}^{2}}}=x\) (actually \(\sqrt{{{x}^{2}}}=\left| x \right|\) since * x* can be negative) since \(x\,\times \,x={{x}^{2}}\). We also learned that taking the square root of a number is the same as raising it to \(\frac{1}{2}\), so \({{x}^{\frac{1}{2}}}=\sqrt{x}\). Also, remember that when we take the square root, there’s an invisible 2 in the radical, like this: \(\sqrt[2]{x}\). Also note that what’s under the radical sign is called the

**radicand**(

**in the previous example), and for the**

*x***nth root**, the

**index**is

**n**(

**2**, in the previous example, since it’s a square root). With a

**negative exponent**, there’s nothing to do with negative numbers! You move the base from the numerator to the denominator (or denominator to numerator) and make it positive! So if you have a base with a negative number that’s not a fraction, put 1 over it and make the exponent positive. And if the negative exponent is on the outside parentheses of a fraction, take the

**reciprocal**of the fraction and make the exponent positive. Some examples: \(\displaystyle {{x}^{-2}}={{\left( \frac{1}{x} \right)}^{2}}\) and \(\displaystyle {{\left( \frac{y}{x} \right)}^{-4}}={{\left( \frac{x}{y} \right)}^{4}}\). Just a note that we’re only dealing with

**real numbers**at this point; later we’ll learn about

**imaginary numbers**, where we can (sort of) take the square root of a negative number.

# Properties of Exponents and Radicals

So remember these **basic rules**:

In algebra, we’ll need to know these and many other basic rules on how to handle exponents and roots when we work with them. Here are the rules/properties with explanations and examples. In the “proof” column, you’ll notice that we’re using many of the algebraic properties that we learned in the **Types of Numbers and Algebraic Properties** section, such as the Associate and Commutative properties. Unless otherwise indicated, **assume numbers under radicals with even roots are positive, and numbers in denominators are nonzero**.

I know this seems like a **lot** to know, but after a lot of practice, they become second nature. You will have to learn the basic properties, but after that, the rest of it will fall in place!

# Putting Exponents and Radicals in the Calculator

We can put exponents and radicals in the graphing calculator, using the carrot sign (**^**) to raise a number to something else, the **square root button** to take the square root, or the **MATH** button to get the cube root or *n*th root. Be careful though, because if there’s not a perfect square root, the calculator will give you a long decimal number that’s not the “**exact value**”. The “**exact value**” would be the answer with the root sign in it! Here are some exponent and radical calculator examples:

# Rationalizing Radicals

Before we work example, let’s talk about **rationalizing radical fractions**. In math, sometimes we have to worry about “proper grammar”. When radicals, it’s improper grammar to have a root on the bottom in a fraction – in the denominator. To fix this, we multiply by a fraction with the bottom radical(s) on both the top and bottom (so the fraction equals 1); this way the bottom radical disappears. Neat trick! Here are some examples:

# Simplifying Exponential Expressions

There are five main things you’ll have to do to simplify exponents and radicals. For the purpose of the examples below, we are assuming that **variables in radicals are non-negative, and denominators are nonzero**.

**get rid of parentheses****()**. Remember that when an exponential expression is raised to another exponent, you multiply exponents. Also remember when you are multiplying numbers and variables and the whole thing is raised to an exponent, you can remove parentheses and “push through” the exponent. Example:**combine bases to combine exponents**. You should add exponents of common bases if you multiplying, and subtract exponents of common bases if you are dividing (you can subtract “up”, or subtract “down” to get the positive exponent as you’ll see). Sometimes you have to match the bases first in order to combine exponents – see last example below. Examples:**get rid of negative exponents**. To get rid of negative exponents, you can simply move a negative exponent in the denominator to the numerator and make it positive, or vice versa. Examples:**simplify any numbers**(like = 2). Also, remember to simplify radicals by taking out any factors of perfect squares (under a square root), cubes (under a cube root), and so on. Example:**combine any like terms**. If you’re adding or subtracting terms with the same numbers (coefficients) and/or variables, you can put these together. Almost think of a**radical expression**(like )**like another variable**. Example: .**Remember that, for the variables, we can divide the exponents inside by the root index – if it goes in exactly, we can take the variable to the outside; if there are any remainders, we have to leave the variables under the root sign.**For example**, ,**since 5 divided by 3 is 1, with 2 left over (for the), and 12 divided by 3 is 4 (for the*x*).*y*

Now let’s put it altogether. Here are some (difficult) examples. Just remember that you have to be really, really careful doing these!

Here are even more examples. **Assume** **variables under radicals are non-negative**.

If we **don’t assume variables under the radicals are non-negative**, we have to be careful with the signs and include **absolute values for even radicals**. Here’s an example:

For all these examples, see how we’re doing the same steps over and over again – just with different problems? If you don’t get them at first, don’t worry; just try to go over them again. You’ll get it! And don’t forget that there are many ways to arrive at the same answers!

# Solving Exponential and Radical Equations

(We’ll see more of these types of problems **here** in the **Solving Radical Equations and Inequalities** section). Now that we know about exponents and roots with variables, we can solve equations that involve them. **The trick is to get rid of the exponents, we need to take radicals of both sides, and to get rid of radicals, we need to raise both sides of the equation to that power.** You have to be a little careful, especially with **even exponents and roots** (the “**evil evens**”), and also when the even exponents are on the **top** of a fractional exponent (this will become the root part when we solve). When we solve for variables with **even exponents**, we most likely will get **multiple solutions**, since when we square positive or negative numbers, we get positive numbers. Also, all the answers we get may not work, since we **can’t take the even roots of negative numbers**. **So it’s a good idea to always check our answers when we solve for roots (especially even roots)!** Let’s first try some equations with **odd exponents and roots**, since these are a little more straightforward. (Notice when we have **fractional** exponents, the radical is still **odd** when the **numerator **is odd). Now let’s solve equations with **even roots**. Note that when we take the even root (like the square root) of both sides, we have to include the **positive** and the **negative** solutions of the roots. (Notice when we have **fractional** exponents, the radical is still **even** when the **numerator **is even.) Again, when the original problem contains an even root sign, we need to check our answers to make sure we have end up with** no negative numbers under the even root sign** (no negative radicands). Also, if we **have ****squared both sides** (or raised both sides to an even exponent), we need to **check our answers to see if they work**. The solutions that don’t work when you put them back in the original equation are called **extraneous solutions**. Again, we’ll see more of these types of problems in the **Solving Radical Equations and Inequalities** section.

And here’s one more where we’re solving for one variable in terms of the other variables:

**Note: **You can also check your answers using a **graphing calculator** by putting in what’s on the left of the = sign in “” and what’s to the right of the equal sign in “”. You can then use the intersection feature to find the solution(s); the solution(s) will be what ** x** is at that point. Here are those instructions again, using an example from above:

# Solving Simple Radical Inequalities

Note again that we’ll see more problems like these, including how to use sign charts with solving radical inequalities here in the **Solving Radical Equations and Inequalities **section. Just like we had to solve linear inequalities, we also have to learn how to solve inequalities that involve exponents and radicals (roots). We’ll do this pretty much the same way, but again, we need to be careful with **multiplying and dividing by anything negative**, where we have to change **the direction of the inequality sign**. We also have to be careful that our answer still keeps what’s **under an even radical** to be **positive**, so we have to create another inequality and set what’s under the even radical to greater than or equal to 0, solve for ** x**, and take the intersection of both solutions. The reason we take the intersection of the two solutions is because

**both**must work. With

**odd roots**, we don’t have to worry – we just raise each side that power, and solve! Here are some examples:

**Learn these rules, and practice, practice, practice!**

**For Practice**: Use the **Mathway** widget below to try an** Exponent **problem. Click on **Submit** (the blue arrow to the right of the problem) to see the answer.

You can also type in your own problem, or click on the three dots in the upper right hand corner and click on “Examples” to drill down by topic.

If you click on **Tap to view steps**, or **Click Here**, you can register at **Mathway** for a **free trial**, and then upgrade to a paid subscription at any time (to get any type of math problem solved!).

On to **Introduction to Multiplying Polynomials** – you are ready!

I came looking for help solving for x in an equation where all the instances of x aren’t roots (and the one that is is in denominator):

1+1/2 x^(-1/2)-2x

THX

I’m not sure exactly how to parse your expression, but you’ll probably have to rationalize the fraction with x^1/2 and then use common denominators to add the three expressions. I haven’t gotten into rationals yet (variables in the denominator) – it will be later. But my best guess for your problem would be 1 + 1/(2sqrtx) – 2x = 1 + sqrt(x)/(2x) – 2x = (2x + sqrt(x) – 4x^2)/2x. Does that make sense?

Fourth root of 64 is 4? Isn’t rather 2√4 ?

Yes – thanks so much for pointing that out. I will change it 🙂

How would you solve the 8th root of y squared ??

The 8th root of y squared would be (y^2)^(1/8) or y^(1/4) or y raised to the 1/4, or the 4th root of y. Remember that when you raise something to something and raise it again, you multiply the exponents. For example (16^(1/4))^2 = 16^(1/4 x 2) = 16^(1/2) = square root of 16 = 4. Does that make sense?

I need help trying to solve a problem.

cube root √64 a^9 b^7 c

4 a^3 b^2 [b] [] is absolute value

How do you or can you solve

Negative 4th root of 256? Show all steps if It can be solved.

Thanks for asking; I will cover this in the Pre-Calc Trig sections. You need Trig/Complex numbers to solve it. 🙂

how would I solve the problem

Combine the like radicals in

^5sqrt(2x) – ^5sqrt(32x^5) + ^5sqrt(-64x) + ^3sqrt(8x^3)

Steve,

Thanks for writing! I’m not sure if I got the original problem correct, but here’s what I got: http://www.shelovesmath.com/algebra/beginning-algebra/solving-algebraic-equations/capture-3/

i am having a horrible amount of trouble with dividing and simplifying radicals. Ex: (cuberoot 10y^8)/(cuberoot 27x^9) This is an entire section in my algebra course. please help!

Thanks for writing! With these types of problems, just see if you can take out any cube roots, and you can! So we’ll be left with y^2cuberoot(10y^2)/(3x^3). Do you see how with numbers, we just take out any cube roots, and with letters (variables), we divide 3 into the exponents, and leave remainders inside the cube roots? Does that make sense? Lisa

i came to try to figure out the square root of 12y^5, but i couldn’t figure it out

Thanks for writing! The square root of 12y^5 would be sqrt(12y^5) = sqrt(12) x y^(5/2) = 2sqrt(3) x y ^ (5/2). Does that make sense? When we take roots, we put the root on the bottom of the fraction of the exponent. Lisa

How would I solve the expression: (2)2=(5t^2)^2? I am having the hardest time trying to calculate this!

Here’s how I would solve: 4 = 25t^4; 4/25 = t^4; t^2= +/- 2/5; t = +/- sqrt(2/5), or +/- sqrt(10)/5, if you rationalize. Does that make sense? Lisa

Okay that’s the answer I keep getting. I think I’m missing something from the expression. Thank you so much for your quick response!

Why does sqrt(x^2-v^2) equal x*sqrt(1-y^2/x^2)?

Thanks for writing! Sqrt(x^2 – y^2) = sqrt(x^2(1 – y^2/x^2)) (multiply it back through to see why!), which is sqrt(x^2)*sqrt(1 – y^2/x^2) = x*(1 – y^2/x^2). Does that make sense? Lisa

I think I get it now, thanks.

So basically, taking the x^2 factor out, I’m dividing both the x^2 and y^2 factors in the brackets by x^2, giving me 1-y^2/x^2 inside the radical, and I need to multiply this (I mean sqrt(1-y^2/x^2)) by sqrt(x^2) (which is x), obviously in order to obtain the same result.

Now you’ve gone through it, it makes complete sense, and I’m kicking myself that I didn’t understand it. Thank you very much for the answer 😉

I have created a video solution for this problem. I think it’s a good example for working with radicals. The video will appear on my website, this afternoon. (www.PaulkUSA.com)

The video solution will be posted on:

http://www.PaulkUSA.com

Math Videos, Other Math Videos

Algebra Radical Example 01

I have another question for you, and I hope that I can ask it in a way that makes sense 😉

If a^2 = b^2 + c^2, then a does not equal b + c.

I know this to be true, because I can demonstrate it geometrically with triangles, and numerically just by putting numbers in and seeing what happens – kind of trial and error.

For instance 5^2 = 3^ + 4^2, but 5 doesn’t equal 3 + 4.

What I’m looking for, though, is the algebraic proof or law that explains why, if a^2 = b^2 + c^2, then it is impossible for it to also be true that a = b + c.

In essence that the square root of the sum of 2 squares cannot equal the sum of the individual square roots of those 2 squares.

It’s something I know to be true, but I just don’t know how that would look written as a proof – as some algebraic equation which can be shown to be true in all circumstances.

If this question is nonsensical, though, could you tell me why?

Though I have a feeling you’ll again explain it, it will again be incredibly simple, and I’ll again kick myself for trying to do maths after half a bottle of wine.

Thank you again for your time 😉

Thanks for writing again. Here’s how I’d “prove” it: Let a + b = c (I’m trying to prove it a^2 + b^2 doesn’t equal c^2). Then a = c – b. So substitute to show that a^2 + b^2 doesn’t equal c^2: (c – b)^2 + b^2 = c^2 – 2cb + b^2 +b^2. If this equal c^2, then c^2 – 2cb + 2b^2 = c^2, which would mean that 2b^2 = 2cb, which would mean that b would have to equal c. This would work only if a = 0.

Does this make sense? Lisa

Thank you so much.

I had to write it out vertically to get the gist. Though I think you meant c^2 – 2cb + 2b^2 = c^2, instead of c^2 – 2cb + b^2 = c^2, am I correct? It seemed to agree with your statement that it would then require 2bc to equal 2b^2, and this for b to equal c , and thus to make sense that this is only possible if a is 0 – thus proving that it is both only possible if a = 0 and if impossible if all three terms represent different integer values.

Thank you again – your impeccable guidance has really helped 😉

Yes, you are right – thanks for correcting that! Lisa

I’d like to provide another algebraic proof, similar to Lisa’s proof.

Problem: If a^2 = b^2 +c^2 Prove: a is not = b+c

Solution: Let a = b+c

If we substitute this into the original equation and find it false,

then a cannot = b+c.

Given: a^2 = b^2 + c^2

(b+c)^2 = b^2 + c^2

b^2 + 2bc + c^2 = b^2 + c^2 Only true if b or c = 0

(Cd)^2/3(a^1/4d)^3 in radical forma

Is it a to the 1/4 times d, or a to the 1 over 4d. Thanks 😉 Lisa

Can someone please solve these two equations?

Simplify the expression. Do not assume the variables represent positive numbers.

sqrt(45 a^3 b^2)=

Rationalize the denominator in the following expression.

2/root(3)(7)

Thanks for writing!! sqrt(45a^3b^2) = 3abs(a)abs(b)sqrt(5a). Does that make sense? If variables can be negative, when we take an even root of them, we take the positive number out; thus, the absolute values.

For the second one, I’m not sure if the 7 is under the root, but I did the problem like it’s not. So I multiplied by sqrt(3)/sqrt(3) to get 2sqrt(3)/21. Does that make sense? Lisa

I got the same answer for the first problem, but I interpreted the second problem differently so I got a different answer. I took root(3) to mean the 3rd root. Usually, sqrt means the square root or root(2). So, for the second problem, I assumed we started with: 2 over the third root of 7.

I multiplied the numerator and denominator by 7^(2/3) to get my answer:

2*root(3)(49)/7

I will create a video, called “Radical Example 02” on my website. Check it out!

Go to: http://www.paulkUSA.com Then, click on Math Videos

Hello Lisa

Reference to your problem (fourthroot of 64a^7b^8).”Assume variables under radical are non negative”

Your reducing the fmourth root of a^7 to ( a (fourth root of a^3) got me to thinking about finding the lelnth

root of any x^m root. I figured out a rather easy method of finding the any nth root of any a^m. While I thought this was sheer brilliance on my part, I confess I haven’t a clue how to explain it.

Example; the third root of a^7 = reduces to (a (third root of a^3) a=2, ans: 5.03968

Reducing this further, (a^2 (times third root of a) a=2, ans: 5.03968

Reducing this further (a^3 (times third root of a^(-2)) a=2 ans: 5.03968

Reducing this further (a^4 (times third root of a^(a^(-5)) a=2 ans 5.039698

I can find the nth root of any a^m root. I simply add an exponent to the a, a^2, a^3 etc. multiply it times the a^m and subtract the original exponent. That is, a^4 (3 sqrt = 4 x 3 -7 = sqrt a^(-5).

Could you explain why this is true?

Again I am humbled by my lack of real understanding. I blame these tremendous math programs that solve the most esoteric problems but do not explain, beyond, rrmaybe a few elementary steps. I wonder if you have ever considered compiling a book of your work? It would be an extraordinary contribution.

Thank you Lisa.

Manuel

r

Manuel,

Thanks so much for writing! You found an amazing result – I’m so so impressed. I think what you’re doing actually is subtracting integers (1, 2, 3, etc) from the exponent of a. The nth root of a^m is actually a^(m/n). We can write m/n into 2 fractions, for example, n/n + (m-n)/n, which would produce your first result, where n = 3 and m = 7 (a^1 * a^1/3), and so on. Does that make sense? Try the other numbers and see if it does.

Do you really think I should put all this in a book? I’m thinking about creating a series of e-books – maybe it’s a good idea!

Lisa

Hi,

Did you forget to include the negative possibility for the following problem: (y+2)^(2/3) = 4

y = 6 and Y = -10

thanks

Thank you SO MUCH for finding this error; you are absolutely correct! I have switched the problem to make it with an odd exponent. Thanks again for could you proof read the entire site? lol Lisa

Can u help me solve this problems..with step plss

1.)8th root of 16x^8y^4

2.)6th root of 1/27x^3y^9

3.)1/4th root of 72xy^8

4.)8/5 root of 2 + 2 root of 5

Thanks for writing: Here’s what I’d get: 1) (2^4)^(1/8)*x*(y^4)^(1/8) = sqrt(2)xy^1/2 2) 3^(-1/2)x^(1/2)y^(3/2) 3) 2^(12)*3^(8)x^4y^32 4) 3.213 (had to put in calculator 😉 These are tough – the idea is you want to raise the expression by the reciprocal of the root. Does that make sense? Lisa

Hi! may I ask how to simplify rational exponents in calculus?

Thanks for writing! Could you give me an example? Lisa

what about (1/4^a + 1/3^b)(1/4^a – 1/3^b)

i been stock in the solution

Thanks for writing. I assume you mean the 4^a and 3^a are in the denominator? If so, use difference of squares to get 1/(16a^2) – 1/(9b^2). Does that make sense? Lisa

Can anyone help me with this problem? Express cuberoot 2 and 4th root 3 in the same order.

I’m not sure if this what you’re asking, but it does turn out that the cube root of 2 can be expressed as 16^(1/12) and the 4th root of 3 can be expressed as 27^(1/12). Is that what you mean? Lisa

Q={y/y is an integer and <−5<y−3}

I think the only element would be {-4}? Does this make sense? Lisa