Radical Sign

Published: 6 Aug 2023

Mathematics sometimes feels like a magical world full of interesting symbols and concepts. One such symbol that you might have come across is the radical sign $ ( \sqrt{ \quad } ) $. The radical sign $ ( \sqrt{ \quad } ) $ might look like a simple little symbol, but it holds a significant role in mathematics.
So, embrace the power of radicals, and let them guide you through the captivating landscapes of algebra and beyond.

Table of Content
  1. Radical Sign MCQs

Radical Sign MCQs

1. In $ \sqrt[n]{a},\ then \ \sqrt{ \quad } $ is called
O Index
O Radical
O Radicand
O All of them
Show Answer

Radical
Explanation:


2. In $ \sqrt[n]{a}, \ then \ a $ is called
O Index
O Radical
O Radicand
O All of them
Show Answer

Radicand
Explanation:


3. In $ \sqrt[n]{a}, \ then \ n $ is called
O Index
O Radical
O Radicand
O All of them
Show Answer

Index
Explanation:


4. The exponential form of $ \sqrt[n]{a} $ is
O $ a^n $
O $ a^2 $
O $ a^{\frac{2}{n}} $
O $ a^{\frac{1}{n}} $
Show Answer

$ a^{\frac{1}{n}} $
Explanation:
It is the General Exponential form of any radical form


5. $ \sqrt{2} $ , the index is
O 0
O 1
O 2
O All of them
Show Answer

2
Explanation:
If the index is not given means it is 2 which we cannot write.


6. $ \sqrt{1}= $
O 1
O 0
O $ -1 $
O 2
Show Answer

1
Explanation:
Square root of 1 will always be 1.


7. $ \sqrt[3]{1}= $
O 1
O 0
O $ -1 $
O 2
Show Answer

1
Explanation:
Cube root of 1 will always be 1


8. $ \sqrt[3]{1}= $
O 1
O 0
O $ -1 $
O 2
Show Answer

1
Explanation:
Forth root of 1 will always be 1.


9. $ \sqrt{36}= $
O 6
O 0
O $ -6 $
O 4
Show Answer

6
Explanation:
$ \sqrt{36}=\sqrt{6^2} $
$ \sqrt{36}=(6^2)^\frac{1}{2} $
$ \sqrt{36}=6 $


10. $ \sqrt[3]{216}= $
O 4
O 5
O 6
O 7
Show Answer

6
Explanation:
$ \sqrt[3]{216}=\sqrt[3]{6^3} $
$ \sqrt[3]{216}=(6^3)^\frac{1}{3} $
$ \sqrt[3]{216}=6 $


11. $ \sqrt[4]{256}= $
O 4
O 5
O 6
O 7
Show Answer

4
Explanation:
$ \sqrt[4]{256}=\sqrt[4]{4^4} $
$ \sqrt[4]{256}=(4^4)^\frac{1}{4} $
$ \sqrt[4]{256}=4 $


12. $ \sqrt[4]{625}= $
O 4
O 5
O 6
O 7
Show Answer

5
Explanation:
$ \sqrt[4]{625}=\sqrt[4]{5^4} $
$ \sqrt[4]{625}=(5^4)^\frac{1}{4} $
$ \sqrt[4]{625}=5 $


13. $ \sqrt[4]{1296}= $
O 4
O 5
O 6
O 7
Show Answer

6
Explanation:
$ \sqrt[4]{1296}=\sqrt[4]{6^4} $
$ \sqrt[4]{1296}=(6^4)^\frac{1}{4} $
$ \sqrt[4]{1296}=6 $


14. If $ x^2=16 $ , then x=
O 4
O $ -4 $
O Both a & b
O None of these
Show Answer

Both a & b
Explanation:
This means what numbers squared becomes 16. Thus $ x \ can \ be \ 4 \ or \ -4 \ like \ (4)^2=16 \ and \ also \ (-4)^2=16. $
Hence the value of $ x=\pm 4 $.


15. If $ x=\sqrt{16} $ , then $ x= $
O 4
O $ -4 $
O Both a & b
O None of these
Show Answer

4
Explanation:
Here $ x $ is the principal square root of 16, which has always a positive value such is $ x=4 $.


16. $ \sqrt[4]{1296} $ is called root
O $ 4^{th} $
O $ 5^{th} $
O $ 2^{nd} $
O Square
Show Answer

$ 4^{th} $
Explanation:
Here is the index of 4, thus it is called $ 4^{th} $ root.


17. $ \sqrt[3]{64}= $
O 4
O $ -4 $
O Imaginary
O None of these
Show Answer

4
Explanation:
$ \sqrt[3]{64}=\sqrt[3]{4^3} $
$ \sqrt[3]{64}=(4^3)^\frac{1}{3} $
$ \sqrt[3]{64}=4 $


18. $ \sqrt[3]{-64}= $
O 4
O $ -4 $
O Imaginary
O None of these
Show Answer

$ -4 $
Explanation:
If a is negative, then n must be odd for the nth root of a to be a real number.
$ \sqrt[3]{-64}=\sqrt[3]{(-4)^3} $
$ \sqrt[3]{-64}=\left[(-4)^3\right]^\frac{1}{3} $
$ \sqrt[3]{-64}=-4 $


19. $ \sqrt{-64}= $
O 4
O $ -4 $
O Imaginary
O None of these
Show Answer

Imaginary
Explanation:
If radicand is negative, then index must be odd, here the index is 2 which is even.
Hence, $ \sqrt{-64}= $ imaginary


20. $ \sqrt[n]{0}= $
O 1
O 0
O n
O $ -0 $
Show Answer

0
Explanation:
If $ a $ is zero, then
$ \sqrt[n]{0}=0 $


21. $ \sqrt[n]{ab}= $
O $ \sqrt[n]{a} \cdot \sqrt[n]{b} $
O $ \sqrt[n]{a}+\sqrt[n]{b} $
O $ \sqrt[n]{a} \cdot \sqrt{b} $
O $ \sqrt{a} \cdot \sqrt[n]{b} $
Show Answer

$ \sqrt[n]{a} \cdot \sqrt[n]{b} $
Explanation:
It is the product rule of Radical.


22. $ \sqrt[n]{\frac{a}{b}}= $
O $ \sqrt[n]{a} \cdot \sqrt[n]{b} $
O $ \sqrt[n]{a}+\sqrt[n]{b} $
O $ \sqrt[n]{a}-\sqrt[n]{b} $
O $ \frac{\sqrt[n]{a}}{\sqrt[n]{b}} $
Show Answer

$ \frac{\sqrt[n]{a}}{\sqrt[n]{b}} $
Explanation:
It is the Quotient rule of Radicand.


23. $ 2 \sqrt{\frac{150 x y}{3 x}}= $
O $ 2 \sqrt{y} $
O $ 10 \sqrt{y} $
O $ 2 \sqrt{2 y} $
O $ 10 \sqrt{2 y} $
Show Answer

$ 10 \sqrt{2 y} $
Explanation:
$ 2 \sqrt{\frac{150 x y}{3 x}}=2 \sqrt{50y} $
$ 2 \sqrt{\frac{150 x y}{3 x}}=2 \sqrt{25 \times 2y} $
$ 2 \sqrt{\frac{150 x y}{3 x}}=2 \times 5 \sqrt{2y} $
$ 2 \sqrt{\frac{150 x y}{3 x}}=10 \sqrt{2y} $


24. $ \sqrt[n]{a}= $
O $ a^{\frac{m}{n}} $
O $ a^{\frac{1}{n}} $
O $ a^{\frac{n}{m}} $
O All of them
Show Answer

$ a^{\frac{1}{n}} $
Explanation:
It is the exponential form of radical.


25. $ a^{\frac{m}{n}}= $
O $ \sqrt[n]{a} $
O $ \sqrt[n]{a^m} $
O Both a & b
O None of these
Show Answer

$ \sqrt[n]{a^m} $
Explanation:
$ a^{\frac{m}{n}}=(a^m)^\frac{1}{n} $
$ a^{\frac{m}{n}}=\sqrt[n]{a^m} $


26. $ \sqrt{13} $ is ________ form
O Exponential
O Radical
O Quadratic
O Cubic
Show Answer

Radical
Explanation:
This is the way to represent the radical form.


27. $ 13^2 $ is ________ form.
O Exponential
O Radical
O Quadratic
O Cubic
Show Answer

Exponential
Explanation:
This is the way to represent the Exponential form.


28. $ 2^4= $
O 16
O $ -16 $
O Both a & b
O None of these
Show Answer

16
Explanation:
$ 2^4= 2 \times 2 \times 2 \times 2 $
$ 2^4= 16 $


29. $ -2^4= $
O 16
O $ -16 $
O Both a & b
O None of these
Show Answer

$ -16 $
Explanation:
$ -2^4= -(2 \times 2 \times 2 \times 2) $
$ -2^4= -16 $


30. $ (-2)^4= $
O 16
O $ -16 $
O Both a & b
O None of these
Show Answer

16
Explanation:
$ (-2)^4= -2 \times -2 \times -2 \times -2 $
$ (-2)^4= 16 $


31. $ \sqrt[n]{a^m}= $
O $ a^{\frac{m}{n}} $
O $ a^{\frac{1}{n}} $
O $ a^{\frac{n}{m}} $
O All of them
Show Answer

$ a^{\frac{m}{n}} $
Explanation:
$ \sqrt[n]{a^m}=(a^m)^\frac{1}{n} $
$ \sqrt[n]{a^m}= a^{\frac{m}{n}} $

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Jawad Khan
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