Roots of Real Numbers and Radicals
Questions with Solutions
Grade 10 questions on roots of numbers and radicals with solutions are presented.
Definition
\[
\large{\textcolor{red}{x \text{ is the } n^\text{th} \text{ root of a number } y \text { is equivalent to } x^n = y.}}
\]
For \( \large {\textcolor{red} {n = 2} } \), the \( n^\text{th} \) root is called the \( \large {\textcolor{red} {\text{square root}}} \).
For \( \large {\textcolor{red} {n = 3} } \), the \( n^\text{th} \) root is called the \( \large {\textcolor{red} {\text{cubic root}}} \).
Examples
1) Since \( 3^2 = 9 \), \( 3 \) is the square (\( n = 2 \)) root of \( 9 \).
2) Since \( (-3)^2 = 9 \), \( -3 \) is also a square root of \( 9 \).
3) Since \( (-2)^3 = -8 \), \( -2 \) is the cubic (\( n = 3 \)) root of \( -8 \).
4) Since \( 3^4 = 81 \) and \( (-3)^4 = 81 \), the fourth roots of \( 81 \) are \( 3 \) and \( -3 \).
Properties of Roots of Real Numbers
1) For \( n \) even and \( y \) positive, there are two \( n^\text{th} \) roots of y
Example
Since 104=10000 and (-10)4 = 10000, the fourth roots of 10000 are 10 and -10.
2) For \( n \) even and \( y \lt 0 \), there are no real \( n^\text{th} \) roots of \( y \)..
Example
The square root of \(-4\) is not a real number since no real number \(x\) exists such that \(x^2 = -4\).
The fourth root of \(-16\) is not a real number since no real number \(x\) exists such that \(x^4 = -16\).
3) For\( n \) odd, there is always one \( n^\text{th} \) root of \( y \).
Example
The cubic \( (n=3) \) root of \( 8 \) is equal to \( 2 \).
The fifth root \( (n=5) \) of \(-100000\) is equal to \( -10 \)
Principal Root
For \( n \) even, the principal root is the positive root. For \( n \) odd there is only one root and it is the principal root.
Examples
The principal \( 6^\text{th} \) of \( 64 \) is equal to \( 2 \) because \( 2^6 = 64 \).
The principal cubic root of \( -64 \) is equal to \( - 4 \) because \( (-4)^3 = - 64 \).
Radical Notation
The symbol \( \sqrt{\hphantom{9}} \) is called a radical and is used to indicate the principal root of a number as follows:
\[
\large{\sqrt[n]{y}}
\]
where \( n \) is called the index of the radical and \( y \) is called the radicand.
Examples
\[
\sqrt[6]{64} = 2
\]
\[
\sqrt[3]{-27} = -3
\]
Because of its widespread use, the square root \( (n=2) \) of \( y \) is written as \( \sqrt{y} \) without indicating the index.
Questions With Solutions
- What is (are) the \( 4^{th} \) root(s) of 16?
- What is (are) the \( 7^{th} \) root(s) of \(-1\)?
- Which number has a fifth root equal to \(-3\)?
- If the sixth root of \( y \) is equal to \(-5\), then \( y = \underline{\hspace{2cm}} \)
- What is (are) the \( 20^{th} \) root(s) of \(-1\)?
- What is the principal \( 4^{th} \) root of 81?
- \[ \sqrt{-4} = \underline{\hspace{2cm}} \]
- \[ \sqrt[10]{\dfrac{10}{-10}} = \underline{\hspace{2cm}} \]
- \[ \sqrt[3]{-1} = \underline{\hspace{2cm}} \]
- \[ \sqrt[3]{1000} = \underline{\hspace{2cm}} \]
- \[ \sqrt[5]{\dfrac{64}{2}} = \underline{\hspace{2cm}} \]
- \[ \sqrt{(-23)^2} = \underline{\hspace{2cm}} \]
- \[ \sqrt{4^6} = \underline{\hspace{2cm}} \]
- \[ \sqrt[7]{5^7} = \underline{\hspace{2cm}} \]
- \[ \sqrt[4]{10^2 - 6^2} = \underline{\hspace{2cm}} \]
- \[ \sqrt[3]{2^9} = \underline{\hspace{2cm}} \]
- \[ \sqrt{6 \dfrac{1}{4}} = \underline{\hspace{2cm}} \]
- Use a calculator to approximate the following to 3 decimal places:
- \(\sqrt[3]{4} =\)
- \(\sqrt{1.3} =\)
- \(\sqrt{\dfrac{2}{5}} =\)
- \(\sqrt[3]{2 \dfrac{1}{3}} =\)
Solutions to the Above Problems
- What is (are) the \( 4^{th} \) root(s) of 16?
\(2\) and \(-2\) because \(2^4 = 16\) and \((-2)^4 = 16\).
- What is (are) the \( 7^{th} \) root(s) of \(-1\)?
\(-1\) because \((-1)^7 = -1\).
- Which number has a fifth root equal to \(-3\)?
\((-3)^5 = -243\).
- If the sixth root of \( y \) is equal to \(-5\), then \( y = \underline{\hspace{2cm}} \)
\(y = (-5)^6 = 15625\).
- What is (are) the \( 20^{th} \) root(s) of \(-1\)?
If \(x\) is the \(20^{th}\) 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 \(20^{th}\) root of \(-1\) is not a real number.
- What is the principal \( 4^{th} \) root of 81?
\(81 = 3^4\) and \(81 = (-3)^4\). Hence \(81\) has two fourth roots but the principal one is the positive one \(3\).
- \[ \sqrt{-4} = \text{is not a real number} \]
- \[ \sqrt[10]{\dfrac{10}{-10}} = \sqrt[10]{-1} = \text{is not a real number} \]
- \[ \sqrt[3]{-1} = -1 \]
- \[ \sqrt[3]{1000} = \sqrt[3]{10^3} = 10 \]
- \[ \sqrt[5]{\dfrac{64}{2}} = \sqrt[5]{32} = \sqrt[5]{2^5} = 2 \]
- \[ \sqrt{(-23)^2} = \sqrt{529} = 23 \]
- \[ \sqrt{4^6} = \sqrt{(4^3)^2} = 4^3 = 64 \]
- \[ \sqrt[7]{5^7} = 5 \]
- \[ \sqrt[4]{10^2 - 6^2} = \sqrt[4]{100 - 36} = \sqrt[4]{64} = \sqrt[4]{2^6} = 2^{6/4} = 2^{3/2} = \sqrt{8} = 2\sqrt{2} \approx 2.828 \]
- \[ \sqrt[3]{2^9} = \sqrt[3]{(2^3)^3} = 2^3 = 8 \]
- \[ \sqrt{6 \dfrac{1}{4}} = \sqrt{\dfrac{25}{4}} = \dfrac{5}{2} = 2.5 \]
- Use a calculator to approximate the following to 3 decimal places:
- \(\sqrt[3]{4} \approx 1.587\)
- \(\sqrt{1.3} \approx 1.140\)
- \(\sqrt{\dfrac{2}{5}} = \sqrt{0.4} \approx 0.632\)
- \(\sqrt[3]{2 \dfrac{1}{3}} = \sqrt[3]{\dfrac{7}{3}} \approx 1.326\)
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