Last updated on May 26th, 2025
If a number is multiplied by the same number, the result is a square. The inverse of the square is a square root. The concept of square roots extends into complex numbers for negative values. Here, we will discuss the square root of -48.
The square root of a negative number involves imaginary numbers. In this case, the square root of -48 is expressed in the form of an imaginary number. The square root of -48 can be represented as √(-48) = √(48) * √(-1) = 4√3 * i, where i is the imaginary unit, defined as √(-1).
The square root of a negative number is not a real number, so we use the concept of imaginary numbers. The process involves finding the square root of the positive part and multiplying it by the imaginary unit i.
Step 1: Identify the positive part of -48, which is 48.
Step 2: Find the square root of 48. The prime factorization of 48 is 2 x 2 x 2 x 2 x 3, which can be grouped as (2 x 2) x (2 x 2) x 3 = 4 x 4 x 3.
Step 3: The square root of 48 is 4√3, as 4 is the square root of 16 (which is 4 x 4), and √3 remains under the radical.
Step 4: Include the imaginary unit i, thus √(-48) = 4√3 * i.
To find the square root of -48 using prime factorization, we focus on the positive part 48 and then include the imaginary unit:
Step 1: Find the prime factors of 48, which are 2 x 2 x 2 x 2 x 3.
Step 2: Pair the prime factors to simplify: (2 x 2) x (2 x 2) x 3 = 4 x 4 x 3.
Step 3: The square root of 48 becomes 4√3 since 4 is the square root of 16.
Step 4: Since we are dealing with -48, we multiply by i, hence √(-48) = 4√3 * i.
The long division method is typically used for non-perfect square numbers. However, when dealing with negative numbers, the focus is on the positive component:
Step 1: Start by considering the positive part, 48. Group as 48.
Step 2: Find the closest perfect square less than 48, which is 36. The square root of 36 is 6.
Step 3: Approximate further to find that √48 is approximately 6.9282.
Step 4: Combine with the imaginary unit i, so √(-48) = 6.9282 * i, which simplifies to 4√3 * i.
The approximation method involves estimating the square root using nearby perfect squares:
Step 1: The closest perfect squares around 48 are 36 (6 squared) and 49 (7 squared).
Step 2: √48 falls between 6 and 7. Calculate the approximation as follows:
Step 3: Using the formula (Given number - smaller perfect square) ÷ (larger perfect square - smaller perfect square), we find (48 - 36) ÷ (49 - 36) = 12 ÷ 13 ≈ 0.923.
Step 4: Add this to the smaller square root, 6 + 0.923 ≈ 6.923, and multiply by i, so √(-48) ≈ 6.923 * i = 4√3 * i.
Students often make errors when dealing with negative square roots, such as forgetting the role of the imaginary unit, i. Let's explore common mistakes and how to avoid them:
Can you help Max find the area of a square box if its side length is given as √(-48)?
The area of the square is -48 square units.
The area of the square = side^2.
The side length is given as √(-48) = 4√3 * i.
Area of the square = (4√3 * i)^2 = -48.
Therefore, the area of the square box is -48 square units.
A square-shaped building measuring -48 square feet is built; if each of the sides is √(-48), what will be the square feet of half of the building?
-24 square feet
The area of the building is -48 square feet.
Half of this area is -24 square feet.
So half of the building measures -24 square feet.
Calculate 2 * √(-48).
8√3 * i
The first step is to find the square root of -48, which is 4√3 * i.
The second step is to multiply 4√3 * i by 2.
So, 2 * (4√3 * i) = 8√3 * i.
What will be the square root of (-24) * 2?
The square root is 4√3 * i
To find the square root, calculate (-24) * 2 = -48.
Then, √(-48) = 4√3 * i.
Therefore, the square root of (-24) * 2 is 4√3 * i.
Find the perimeter of the rectangle if its length ‘l’ is √(-48) units and the width ‘w’ is 6 units.
The perimeter of the rectangle is not a real number.
Perimeter of the rectangle = 2 × (length + width).
Length = 4√3 * i,
Width = 6.
Perimeter = 2 × (4√3 * i + 6), which involves adding an imaginary number to a real number, thus it is not a real number.
Jaskaran Singh Saluja is a math wizard with nearly three years of experience as a math teacher. His expertise is in algebra, so he can make algebra classes interesting by turning tricky equations into simple puzzles.
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