Question

How do you simplify radical expressions ?

Answer 1

Hint: Radical expressions or exponential expressions or surds. These all convey the same meaning. All these are of the form $\sqrt[n]{a}$ where $a$ can be a variable or a constant and $n$ is the it’s power . ${{a}^{\dfrac{1}{n}}}$ also represents surds or radical expressions. Here n can be positive or negative. For example if $n$ is 2 in $\sqrt[n]{a}$ , then that is $\sqrt[2]{a}$ which is nothing but ${{a}^{\dfrac{1}{2}}}$ . We read it as the square root of a. $\sqrt[2]{a}$ can also just be written as $\sqrt{a}$ as both represent the same. Solving a radical expression now deals whether the number inside the root is a composite number or a prime number.

Complete step-by-step solution:
Prime number is a number whose factors are 1 and itself. Example : 2,3,5 etc.
Composite numbers are those whose factors are more than 1 and itself. Example : 10,15,20 etc
For example square root of 4 :
  $ \Rightarrow \sqrt[2]{4}={{4}^{\dfrac{1}{2}}} $
 $ \Rightarrow \sqrt[2]{4}={{\left( {{2}^{2}} \right)}^{\dfrac{1}{2}}} $
$ \Rightarrow \sqrt[2]{4}=\left( {{2}^1} \right) $
$ \Rightarrow \sqrt[2]{4}=2 $
In the first example , we solved a radical expression which has a composite number . Now , let’s solve a radical expression which has a prime number in it.
Square root of 11 :
$\Rightarrow 11=11\times 1$ . The factor of 11 is 1 and itself.
$\Rightarrow \sqrt[2]{11}=\sqrt{11}$ . This can’t be further solved.
Now let’s solve a radical expression of power 3. So we solved the cube root.
Cube root of 8:
  $ \Rightarrow \sqrt[3]{8}={{8}^{\dfrac{1}{3}}} $
 $ \Rightarrow \sqrt[3]{8}={{\left( {{2}^{3}} \right)}^{\dfrac{1}{3}}} $
 $ \Rightarrow \sqrt[3]{8}=\left( {{2}^1} \right) $
 $ \Rightarrow \sqrt[3]{8}=2 $
Cube root of 16 :
\[\begin{align}
  & \Rightarrow \sqrt[3]{16}={{16}^{\dfrac{1}{3}}} \\
 & \Rightarrow \sqrt[3]{16}={{\left( 2\times 2\times 2\times 2 \right)}^{\dfrac{1}{3}}} \\
 & \Rightarrow \sqrt[3]{16}={{\left( {{2}^{4}} \right)}^{\dfrac{1}{3}}} \\
 & \Rightarrow \sqrt[3]{16}={{\left( {{2}^{3}} \right)}^{\dfrac{1}{3}}}\times {{\left( 2 \right)}^{\dfrac{1}{3}}} \\
 & \Rightarrow \sqrt[3]{16}=2\times \sqrt[3]{2}={{2}^{\dfrac{4}{3}}} \\
\end{align}\]
It is better to just leave the answer in the root form rather than finding out it’s decimal form. As few decimal forms are non-terminating non-repeating and non-terminating repeating.
Now let’s try solving radical expressions using variables.
${{4}^{th}}$ root of \[81{{x}^{8}}{{y}^{2}}{{z}^{4}}\]
\[\begin{align}
  & \Rightarrow \sqrt[4]{81{{x}^{8}}{{y}^{2}}{{z}^{4}}}={{\left( 81 \right)}^{\dfrac{1}{4}}}{{\left( {{x}^{8}} \right)}^{\dfrac{1}{4}}}{{\left( {{y}^{2}} \right)}^{\dfrac{1}{4}}}{{\left( {{z}^{4}} \right)}^{\dfrac{1}{\begin{smallmatrix}
 4 \\
\end{smallmatrix}}}} \\
 & \Rightarrow \sqrt[4]{81{{x}^{8}}{{y}^{2}}{{z}^{4}}}={{\left( {{3}^{4}} \right)}^{\dfrac{1}{4}}}\left( {{x}^{2}} \right)\left( {{y}^{\dfrac{1}{2}}} \right)\left( z \right) \\
 & \Rightarrow \sqrt[4]{81{{x}^{8}}{{y}^{2}}{{z}^{4}}}=3z{{x}^{2}}\sqrt{y} \\
\end{align}\]
One with a negative root :
$\begin{align}
  & \Rightarrow {{\left( 125 \right)}^{\dfrac{-1}{3}}}=\dfrac{1}{{{\left( 125 \right)}^{\dfrac{1}{3}}}} \\
 & \Rightarrow {{\left( 125 \right)}^{\dfrac{-1}{3}}}=\dfrac{1}{{{\left( {{5}^{3}} \right)}^{\dfrac{1}{3}}}} \\
 & \Rightarrow {{\left( 125 \right)}^{\dfrac{-1}{3}}}=\dfrac{1}{5} \\
\end{align}$
$\therefore {{a}^{\dfrac{-1}{n}}}=\dfrac{1}{{{a}^{n}}}$.
In this way we solve either variables or constants or functions with both variables and constants.

Note: We have to be careful with the powers and do the sum accordingly. As I have already mentioned , it is better to leave the answers in the root form rather than the decimal form . For example $\sqrt{2}=1.41421356............$ and it keeps going as it is a non-terminating non-repeating decimal and we wouldn’t know where to end it. So just leave it as $\sqrt{2}$ .