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### Section 1.3 : Radicals

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Evaluate the following.

- \(\sqrt[3]{{125}}\)

Show SolutionIn order to evaluate radicals all that you need to remember is

\[y = \sqrt[n]{x}{\mbox{ is equivalent to }}x = {y^n}\]In other words, when evaluating \(\sqrt[n]{x}\) we are looking for the value, \(y\), that we raise to the \(n\) to get \(x\). So, for this problem we’ve got the following.

\[\sqrt[3]{{125}} = 5\hspace{0.25in}{\rm{because }}\hspace{0.25in}{5^3} = 125\] - \(\sqrt[6]{{64}}\)

Show Solution\[\sqrt[6]{{64}} = 2\hspace{0.25in}{\rm{because }}\hspace{0.25in}{2^6} = 64\] - \(\sqrt[5]{{ - 243}}\)

Show Solution\[(\sqrt[5]{{ - 243}} = - 3\hspace{0.25in}{\rm{because }}\hspace{0.25in}{\left( { - 3} \right)^5} = - 243\] - \(\sqrt[2]{{100}}\)

Show Solution\[\sqrt[2]{{100}} = \sqrt {100} = 10\hspace{0.25in}{\mbox{Remember that }}\sqrt[2]{x} = \sqrt x \] - \(\sqrt[4]{{ - 16}}\)

Show Solution\[\sqrt[4]{{ - 16}} = {\rm{n/a}}\]Technically, the answer to this problem is a complex number, but in most calculus classes, including mine, complex numbers are not dealt with. There is also the fact that it’s beyond the scope of this review to go into the details of getting a complex answer to this problem.

Convert each of the following to exponential form.

- \(\sqrt {7x} \)

Show SolutionTo convert radicals to exponential form you need to remember the following formula

\[\sqrt[n]{p} = {p^{\frac{1}{n}}}\]For this problem we’ve got.

\[\sqrt {7x} = \sqrt[2]{{7x}} = {\left( {7x} \right)^{\frac{1}{2}}}\]There are a couple of things to note with this one. Remember \(\sqrt p = \sqrt[2]{p}\) and notice the parenthesis. These are required since both the 7 and the \(x\) was under the radical so both must also be raised to the power. The biggest mistake made here is to convert this as

\[7{x^{\frac{1}{2}}}\]however this is incorrect because

\[7{x^{\frac{1}{2}}} = 7\sqrt x \]Again, be careful with the parenthesis

- \(\sqrt[5]{{{x^2}}}\)

Show Solution\[\sqrt[5]{{{x^2}}} = {\left( {{x^2}} \right)^{\frac{1}{5}}} = {x^{\frac{2}{5}}}\]Note that I combined exponents here. You will always want to do this.

- \(\sqrt[3]{{4x + 8}}\)

Show Solution\[\sqrt[3]{{4x + 8}} = {\left( {4x + 8} \right)^{\frac{1}{3}}}\]You CANNOT simplify further to \({\left( {4x} \right)^{\frac{1}{3}}} + {\left( 8 \right)^{\frac{1}{3}}}\) so don’t do that!!!! Remember that \({\left( {a + b} \right)^n} \ne {a^n} + {b^n}\)!!!!

Simplify each of the following.

- \(\sqrt[3]{{16{x^6}{y^{13}}}}\) Assume that \(x \ge 0\) and \(y \ge 0\) for this problem.

Show SolutionThe property to use here is

\[\sqrt[n]{{xy}} = \sqrt[n]{x}\,\,\sqrt[n]{y}\]A similar property for quotients is

\[\sqrt[n]{{\frac{x}{y}}} = \frac{{\sqrt[n]{x}}}{{\sqrt[n]{y}}}\]Both of these properties require that at least one of the following is true \(x \ge 0\) and/or \(y \ge 0\). To see why this is the case consider the following example

\[4 = \sqrt {16} = \sqrt {\left( { - 4} \right)\left( { - 4} \right)} \ne \sqrt { - 4} \sqrt { - 4} = \left( {2i} \right)\left( {2i} \right) = 4{i^2} = - 4\]If we try to use the property when both are negative numbers we get an incorrect answer. If you don’t know or recall complex numbers you can ignore this example.

The property will hold if one is negative and the other is positive, but you can’t have both negative.

I’ll also need the following property for this problem.

\[\sqrt[n]{{{x^n}}} = x\hspace{0.25in}{\mbox{provided \(n\) is odd}}\]In the next example I’ll deal with \(n\) even.

Now, on to the solution to this example. I’ll first rewrite the stuff under the radical a little then use both of the properties that I’ve given here.

\begin{align*}\sqrt[3]{{16{x^6}{y^{13}}}} & = \sqrt[3]{{8{x^3}{x^3}\,{y^3}{y^3}{y^3}{y^3}\,\,2y}}\\ & = \sqrt[3]{8}\sqrt[3]{{{x^3}}}\sqrt[3]{{{x^3}}}\sqrt[3]{{{y^3}}}\sqrt[3]{{{y^3}}}\sqrt[3]{{{y^3}}}\sqrt[3]{{{y^3}}}\sqrt[3]{{2y}}\\ & = 2x\,x\,y\,y\,y\,y\,\sqrt[3]{{2y}}\\ & = 2{x^2}{y^4}\sqrt[3]{{2y}}\end{align*}So, all that I did was break up everything into terms that are perfect cubes and terms that weren’t perfect cubes. I then used the property that allowed me to break up a product under the radical. Once this was done I simplified each perfect cube and did a little combining.

- \(\sqrt[4]{{16{x^8}{y^{15}}}}\)

Show SolutionI did not include the restriction that \(x \ge 0\) and \(y \ge 0\) in this problem so we’re going to have to be a little more careful here. To do this problem we will need the following property.

\[\sqrt[n]{{{x^n}}} = \left| x \right|\hspace{0.25in}{\mbox{provided \(n\) is even}}\]To see why the absolute values are required consider \(\sqrt 4 \). When evaluating this we are really asking what number did we square to get four? The problem is there are in fact two answer to this : 2 and -2! When evaluating square roots (or any even root for that matter) we want a predicable answer. We don’t want to have to sit down each and every time and decide whether we want the positive or negative number. Therefore, by putting the absolute value bars on the \(x\) we will guarantee that the answer is always positive and hence predictable.

So, what do we do if we know that we want the negative number? Simple. We add a minus sign in front of the square root as follows

\[ - \sqrt 4 = - \left( 2 \right) = - 2\]This gives the negative number that we wanted and doesn’t violate the rule that square root always return the positive number!

Okay, let’s finally do this problem. For the most part it works the same as the previous one did, we just have to be careful with the absolute value bars.

\begin{align*}\sqrt[4]{{16{x^8}{y^{15}}}} & = \sqrt[4]{{16{x^4}{x^4}\,{y^4}{y^4}{y^4}\,{y^3}}}\\ & = 2\left| x \right|\,\left| x \right|\left| y \right|\left| y \right|\left| y \right|\sqrt[4]{{{y^3}}}\\ & = 2{\left| x \right|^2}{\left| y \right|^3}\sqrt[4]{{{y^3}}}\\ & = 2{x^2}{\left| y \right|^3}\sqrt[4]{{{y^3}}}\end{align*}Note that I could drop the absolute value on the \(\left| {{x^2}} \right|\) term because the power of 2 will give a positive answer for \({x^2}\) regardless of the sign of \(x\). They do need to stay on the \(y\) term however because of the power.