Square Root of a Negative Number

There is no root of a negative number since any positive number raised to the second power will result in a positive number.

Suggested Topics to Practice in Advance

  1. Exponents and Roots - Basic
  2. Exponents and Exponent rules
  3. Basis of a power
  4. The exponent of a power
  5. Powers

Practice Square Root of a Negative Number

Examples with solutions for Square Root of a Negative Number

Exercise #1

Choose the largest value

Video Solution

Step-by-Step Solution

Let's begin by calculating the numerical value of each of the roots in the given options:

25=516=49=3 \sqrt{25}=5\\ \sqrt{16}=4\\ \sqrt{9}=3\\ We can determine that:

5>4>3>1 Therefore, the correct answer is option A

Answer

25 \sqrt{25}

Exercise #2

49= \sqrt{49}=

Video Solution

Step-by-Step Solution

To solve this problem, we follow these steps:

  • Step 1: Understand that finding the square root of a number means determining what number, when multiplied by itself, equals the original number.
  • Step 2: Identify the numbers that could potentially be the square root of 4949. These are ±7 \pm7, but by convention, the square root function typically refers to the non-negative root.
  • Step 3: Calculate 7×7=497 \times 7 = 49. This confirms that 49=7 \sqrt{49} = 7.
  • Step 4: Verify using the problem's multiple-choice answers to ensure 77 is among them, confirming choice number .

Therefore, the solution to the problem 49 \sqrt{49} is 7 7 .

Answer

7

Exercise #3

36= \sqrt{36}=

Video Solution

Step-by-Step Solution

Let's solve the problem step by step:

  • Step 1: Understand what the square root means.
  • The square root of a number nn is a value that, when multiplied by itself, equals nn. This is written as x=nx = \sqrt{n}.

  • Step 2: Apply this definition to the number 3636.
  • We are looking for a number xx such that x2=36x^2 = 36. This translates to finding x=36x = \sqrt{36}.

  • Step 3: Determine the correct number.
  • We know that 6×6=366 \times 6 = 36. Therefore, the principal square root of 3636 is 66.

Thus, the solution to the problem is 36=6 \sqrt{36} = 6 .

Among the given choices, the correct one is: Choice 1: 66.

Answer

6

Exercise #4

64= \sqrt{64}=

Video Solution

Step-by-Step Solution

To solve this problem, we'll follow these steps:

  • Step 1: Recognize what finding a square root means
  • Step 2: List known perfect squares to identify which one results in 64
  • Step 3: Verify the square root by calculation

Now, let's work through each step:
Step 1: To find the square root of 64, we seek a number that, when multiplied by itself, equals 64.
Step 2: Consider the sequence of perfect squares: 12=1 1^2 = 1 , 22=4 2^2 = 4 , 32=9 3^2 = 9 , 42=16 4^2 = 16 , 52=25 5^2 = 25 , 62=36 6^2 = 36 , 72=49 7^2 = 49 , 82=64 8^2 = 64 .
Step 3: We see that 82=64 8^2 = 64 . Therefore, the square root of 64 is 8.

Therefore, the solution to this problem is 8 8 .

Answer

8

Exercise #5

Solve the following exercise:

x2= \sqrt{x^2}=

Video Solution

Step-by-Step Solution

In order to simplify the given expression, we will use two laws of exponents:

a. The definition of root as an exponent:

an=a1n \sqrt[n]{a}=a^{\frac{1}{n}} b. The law of exponents for power of a power:

(am)n=amn (a^m)^n=a^{m\cdot n}

Let's start with converting the square root to an exponent using the law mentioned in a':

x2=(x2)12= \sqrt{x^2}= \\ \downarrow\\ (x^2)^{\frac{1}{2}}= We'll continue using the law of exponents mentioned in b' and perform the exponent operation on the term in parentheses:

(x2)12=x212x1=x (x^2)^{\frac{1}{2}}= \\ x^{2\cdot\frac{1}{2}}\\ x^1=\\ \boxed{x} Therefore, the correct answer is answer a'.

Answer

x x

Exercise #6

441= \sqrt{441}=

Video Solution

Step-by-Step Solution

The root of 441 is 21.

21×21= 21\times21=

21×20+21= 21\times20+21=

420+21=441 420+21=441

Answer

21 21

Exercise #7

x=1 \sqrt{x}=1

X=? X=?

Video Solution

Step-by-Step Solution

To solve this problem, we'll follow these steps:

  • Step 1: Square both sides of the equation.
  • Step 2: Simplify the resulting expression.

Now, let's work through each step:

Step 1: We have the equation x=1 \sqrt{x} = 1 .
Square both sides:

(x)2=12 (\sqrt{x})^2 = 1^2

Step 2: Simplify both sides of the equation.

The left side simplifies to x x , since the square and the square root cancel each other out:

x=1 x = 1

The right side simplifies to 1, so we have:

x=1 x = 1

Therefore, the solution to the problem is x=1 x = 1 .

Answer

1

Exercise #8

x=2 \sqrt{x}=2

Video Solution

Step-by-Step Solution

To solve the problem, follow these steps:

  • Step 1: Begin with the equation x=2\sqrt{x} = 2.
  • Step 2: Square both sides of the equation to eliminate the square root.
  • Step 3: Simplify the resulting equation to find xx.

Now, let's proceed through each step:
Step 1: The given equation is x=2\sqrt{x} = 2.
Step 2: Square both sides: (x)2=22(\sqrt{x})^2 = 2^2.
Step 3: This simplifies to x=4x = 4.

Therefore, the value of xx that satisfies x=2\sqrt{x} = 2 is x=4 x = 4 .

Matching this solution with the provided choices, the correct answer is choice 3, which is 4.

Answer

4

Exercise #9

x=6 \sqrt{x}=6

Video Solution

Step-by-Step Solution

To solve this problem, we will perform the following steps:

  • Step 1: Square both sides of the equation x=6 \sqrt{x} = 6 .
  • Step 2: Simplify the equation to find x x .

Let's carry out each step in detail:

Step 1: Square both sides of the equation:
 (x)2=62\ (\sqrt{x})^2 = 6^2

Step 2: Simplify the equation:
Since (x)2=x(\sqrt{x})^2 = x, we have x=36 x = 36 .

Therefore, the value of x x is 36.

Answer

36

Exercise #10

4= \sqrt{4}=

Video Solution

Step-by-Step Solution

To solve this problem, we'll determine the square root of the number 4.

  • Step 1: Recognize that the square root of a number is asking for a value that, when multiplied by itself, yields the original number. Here, we seek a number yy such that y2=4y^2 = 4.
  • Step 2: Identify that 44 is a perfect square. The numbers 22 and 2-2 both satisfy the equation 22=42^2 = 4 and (2)2=4(-2)^2 = 4.
  • Step 3: We usually consider the principal square root, which is the non-negative version. Thus, 4=2\sqrt{4} = 2.

Therefore, the solution to the problem is 2, which corresponds to the correct choice from the given options.

Answer

2

Exercise #11

9= \sqrt{9}=

Video Solution

Step-by-Step Solution

To solve this problem, we want to find the square root of 9.

Step 1: Recognize that a square root is a number which, when multiplied by itself, equals the original number. Thus, we are seeking a number x x such that x2=9 x^2 = 9 .

Step 2: Note that 9 is a common perfect square: 9=3×3 9 = 3 \times 3 . Therefore, the square root of 9 is the number that, when multiplied by itself, gives 9. This number is 3.

Step 3: Since we are interested in the principal square root, we consider only the non-negative value. Hence, the principal square root of 9 is 3.

Therefore, the solution to the problem is 3 3 .

Answer

3

Exercise #12

16= \sqrt{16}=

Video Solution

Step-by-Step Solution

To determine the square root of 16, follow these steps:

  • Identify that we are looking for the square root of 16, which is a number that, when multiplied by itself, equals 16.
  • Recall the basic property of perfect squares: 4×4=16 4 \times 4 = 16 .
  • Thus, the square root of 16 is 4.

Hence, the solution to the problem is the principal square root, which is 4 4 .

Answer

4

Exercise #13

36= \sqrt{36}=

Video Solution

Step-by-Step Solution

To solve this problem, we'll follow these steps:

  • Step 1: Understand the definition of a square root.
  • Step 2: Identify which integer, when squared, gives 36.
  • Step 3: Verify this integer meets the required condition.
  • Step 4: Choose the correct answer from the given choices.

Now, let's work through each step:
Step 1: A square root of a number is a value that, when multiplied by itself, gives the original number. Here, we want y y such that y2=36 y^2 = 36 .
Step 2: We test integer values to find which one squared equals 36. Testing y=1,2,3,4,5, y = 1, 2, 3, 4, 5, and 6 6 gives:
- 12=1 1^2 = 1
- 22=4 2^2 = 4
- 32=9 3^2 = 9
- 42=16 4^2 = 16
- 52=25 5^2 = 25
- 62=36 6^2 = 36

Step 3: The integer 6 6 satisfies 62=36 6^2 = 36 . Therefore, 36=6 \sqrt{36} = 6 .

Step 4: The correct choice from the given answer choices is 6 (Choice 4).

Hence, the square root of 36 is 6 \mathbf{6} .

Answer

6

Exercise #14

64= \sqrt{64}=

Video Solution

Step-by-Step Solution

To solve this problem, we'll determine the square root of 64, following these steps:

  • Step 1: Identify the number whose square root we need to find. The number given is 64.
  • Step 2: Determine which number, when multiplied by itself, equals 64.
  • Step 3: Recall that 8×8=64 8 \times 8 = 64 .

Now, let's work through each step:

Step 1: We are tasked with finding 64 \sqrt{64} . The problem involves identifying the number which, when squared, results in 64.
Step 2: To find this number, we'll check our knowledge of squares. We know that 8 is a significant integer whose square results in 64.
Step 3: Compute: 8×8=64 8 \times 8 = 64 . Hence, 8 8 meets the requirement.

We find that the solution to the problem is 64=8 \sqrt{64} = 8 .

Answer

8

Exercise #15

100= \sqrt{100}=

Video Solution

Step-by-Step Solution

The task is to find the square root of the number 100. The square root operation seeks a number which, when squared, equals the original number. For any positive integer, if x2=100 x^2 = 100 , then x x should be our answer.

Step 1: Recognize that 100 is a perfect square. This means there exists an integer x x such that x×x=100 x \times x = 100 . Generally, we recall basic squares such as:

  • 12=1 1^2 = 1
  • 22=4 2^2 = 4
  • 32=9 3^2 = 9
  • and so forth, up to 102 10^2

Step 2: Checking integers, we find that:

102=10×10=100 10^2 = 10 \times 10 = 100

Step 3: Confirm the result: Since 10×10=100 10 \times 10 = 100 , then 100=10 \sqrt{100} = 10 .

Step 4: Compare with answer choices. Given that one of the choices is 10, and 100=10 \sqrt{100} = 10 , choice 1 is correct.

Therefore, the square root of 100 is 10.

Answer

10

Topics learned in later sections

  1. What is a square root?