Grade 10 questions on roots of numbers and radicals are presented below. Review the definitions and properties, try the practice questions, and click the arrows to view the step-by-step solutions.
1) For $n$ even and $y$ positive, there are two $n^\text{th}$ roots of $y$.
Example: Since $10^4 = 10000$ and $(-10)^4 = 10000$, the fourth roots of $10000$ are $10$ and $-10$.
2) For $n$ even and $y < 0$, there are no real $n^\text{th}$ roots of $y$.
Examples:
3) For $n$ odd, there is always one $n^\text{th}$ root of $y$.
Examples:
For $n$ even, the principal root is defined as the positive root. For $n$ odd, there is only one root, and it is the principal root.
The symbol $\sqrt{\hphantom{9}}$ is called a radical and is used to indicate the principal root of a number as follows:
where $n$ is called the index of the radical and $y$ is called the radicand.
$$ \sqrt[6]{64} = 2 $$
$$ \sqrt[3]{-27} = -3 $$
Note: Because of its widespread use, the square root ($n=2$) of $y$ is written as $\sqrt{y}$ without indicating the index.
What is (are) the $4^\text{th}$ root(s) of $16$?
$2$ and $-2$ because $2^4 = 16$ and $(-2)^4 = 16$.
What is (are) the $7^\text{th}$ root(s) of $-1$?
$-1$ because $(-1)^7 = -1$.
Which number has a fifth root equal to $-3$?
The number is $(-3)^5 = -243$.
If the sixth root of $y$ is equal to $-5$, then $y = \underline{\hspace{2cm}}$
$y = (-5)^6 = 15625$.
What is (are) the $20^\text{th}$ root(s) of $-1$?
If $x$ is the $20^\text{th}$ root of $-1$, then $x^{20} = -1$. There is no real number $x$ raised to an even power that would give a negative number. The $20^\text{th}$ root of $-1$ is not a real number.
What is the principal $4^\text{th}$ root of $81$?
$81 = 3^4$ and $81 = (-3)^4$. Hence $81$ has two fourth roots, but the principal one is the positive one, $3$.
Evaluate: $\sqrt{-4} = \underline{\hspace{2cm}}$
Is not a real number.
Evaluate: $\sqrt[10]{\dfrac{10}{-10}} = \underline{\hspace{2cm}}$
$$ \sqrt[10]{\dfrac{10}{-10}} = \sqrt[10]{-1} $$
Is not a real number.
Evaluate: $\sqrt[3]{-1} = \underline{\hspace{2cm}}$
$$ \sqrt[3]{-1} = -1 $$
Evaluate: $\sqrt[3]{1000} = \underline{\hspace{2cm}}$
$$ \sqrt[3]{1000} = \sqrt[3]{10^3} = 10 $$
Evaluate: $\sqrt[5]{\dfrac{64}{2}} = \underline{\hspace{2cm}}$
$$ \sqrt[5]{\dfrac{64}{2}} = \sqrt[5]{32} = \sqrt[5]{2^5} = 2 $$
Evaluate: $\sqrt{(-23)^2} = \underline{\hspace{2cm}}$
$$ \sqrt{(-23)^2} = \sqrt{529} = 23 $$
Evaluate: $\sqrt{4^6} = \underline{\hspace{2cm}}$
$$ \sqrt{4^6} = \sqrt{(4^3)^2} = 4^3 = 64 $$
Evaluate: $\sqrt[7]{5^7} = \underline{\hspace{2cm}}$
$$ \sqrt[7]{5^7} = 5 $$
Evaluate: $\sqrt[4]{10^2 - 6^2} = \underline{\hspace{2cm}}$
$$ \sqrt[4]{10^2 - 6^2} = \sqrt[4]{100 - 36} = \sqrt[4]{64} = \sqrt[4]{2^6} $$
This can be simplified by fractional exponents:
$$ = 2^{6/4} = 2^{3/2} = \sqrt{8} = 2\sqrt{2} \approx 2.828 $$
Evaluate: $\sqrt[3]{2^9} = \underline{\hspace{2cm}}$
$$ \sqrt[3]{2^9} = \sqrt[3]{(2^3)^3} = 2^3 = 8 $$
Evaluate: $\sqrt{6 \dfrac{1}{4}} = \underline{\hspace{2cm}}$
Convert the mixed number to an improper fraction:
$$ \sqrt{6 \dfrac{1}{4}} = \sqrt{\dfrac{25}{4}} = \dfrac{5}{2} = 2.5 $$
Use a calculator to approximate the following to 3 decimal places: