Square Root of -136
2026-02-28 13:10 Diff
https://brightchamps.com/en-us/math/algebra/square-root-of-minus-136

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Last updated on August 5, 2025

When a number is multiplied by itself, the result is a square. The inverse of squaring a number is finding its square root. The concept of square roots extends to complex numbers, where negative numbers also have square roots. Here, we will discuss the square root of -136.

What is the Square Root of -136?

The square root is the inverse of the square of a number. In the case of negative numbers, the square root involves imaginary numbers. The square root of -136 is expressed in the form of √-136, which can be rewritten as √136 × i, where i is the imaginary unit. The numerical value of √136 is approximately 11.662. Therefore, the square root of -136 is 11.662i, an imaginary number.

Finding the Square Root of -136

The square root of a negative number involves the imaginary unit 'i'. Common methods to calculate square roots include factoring for perfect squares, long division for non-perfect squares, and approximation. However, for negative numbers, we first compute the square root of the positive counterpart and then multiply by i. Let's examine these methods:

  • Prime factorization method
  • Long division method
  • Approximation method

Square Root of -136 by Prime Factorization Method

Prime factorization involves breaking down a number into its prime components. While -136 is not a perfect square, we can find the prime factorization of 136:

Step 1: Finding the prime factors of 136 Breaking it down, we get 2 × 2 × 2 × 17: 2³ × 17

Step 2: Since -136 is negative, the square root involves 'i'. Thus, √-136 = √136 × i = 2√34 × i

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Square Root of -136 by Long Division Method

The long division method is used for non-perfect squares. Let's use this method to find √136, then multiply by i.

Step 1: Group digits of 136 from right to left: 36 and 1.

Step 2: Find n such that n² ≤ 1. Here, n = 1 as 1 × 1 = 1.

Step 3: Subtract 1 from 1 to get 0, then bring down 36.

Step 4: Add 1 to itself to get 2 as the new divisor.

Step 5: Find 2n × n ≤ 36. Let n = 6, then 26 × 6 = 156; adjust to get 216.

Step 6: Subtract 216 from 360 to get 144.

Step 7: Add decimal and bring down two zeroes to get 14400.

Step 8: Find the new divisor. Continue the division until reaching a close approximation. Result: √136 ≈ 11.662, so √-136 = 11.662i

Square Root of -136 by Approximation Method

The approximation method is used for quick calculations. We find the square root of 136 and multiply by i.

Step 1: Identify the perfect squares around 136: 121 (11²) and 144 (12²).

Step 2: 136 is closer to 144, so √136 ≈ 11.662

Step 3: Therefore, the square root of -136 is 11.662i

Common Mistakes and How to Avoid Them in the Square Root of -136

Students often make errors when dealing with square roots of negative numbers, especially involving imaginary units. Let's explore common mistakes:

Problem 1

Can you help Max find the area of a square box if its side length is given as √-144?

Okay, lets begin

The area of the square is -144 square units.

Explanation

The area of the square = side².

The side length is given as √-144.

Area of the square = side² = (√-144)² = (-12i)² = -144

Therefore, the area of the square box is -144 square units (considering the negative value as per the context).

Well explained 👍

Problem 2

A square-shaped building measuring -136 square feet is built; if each of the sides is √-136, what will be the square feet of half of the building?

Okay, lets begin

-68 square feet

Explanation

Divide the given area by 2 as the building is square-shaped:

Dividing -136 by 2 = -68

So half of the building measures -68 square feet.

Well explained 👍

Problem 3

Calculate √-136 × 5.

Okay, lets begin

58.31i

Explanation

First, find the square root of -136, which is 11.662i.

Then multiply 11.662i by 5.

So 11.662i × 5 = 58.31i

Well explained 👍

Problem 4

What will be the square root of (-136 + 4)?

Okay, lets begin

The square root is 12i

Explanation

To find the square root, calculate (-136 + 4) = -132.

The square root of -132 is √132 × i.

Estimate √132 ≈ 11.49, so the result is approximately 11.49i.

Well explained 👍

Problem 5

Find the perimeter of the rectangle if its length ‘l’ is √-136 units and the width ‘w’ is 38 units.

Okay, lets begin

The perimeter of the rectangle is 2 × (√-136 + 38) = 2 × (11.662i + 38) = 76 + 23.324i units.

Explanation

Perimeter of the rectangle = 2 × (length + width)

Perimeter = 2 × (√-136 + 38) = 2 × (11.662i + 38) = 76 + 23.324i units

Well explained 👍

FAQ on Square Root of -136

1.What is √-136 in its simplest form?

The simplest form of √-136 is 11.662i, where i is the imaginary unit, and it represents the square root of -1.

2.Mention the factors of 136.

Factors of 136 are 1, 2, 4, 8, 17, 34, 68, and 136.

3.Calculate the square of -136.

The square of -136 is (-136) × (-136) = 18496.

4.What is an imaginary number?

An imaginary number is a complex number that can be written as a real number multiplied by the imaginary unit 'i', where i² = -1.

5.Is -136 a perfect square?

No, -136 is not a perfect square because it is negative, and perfect squares are non-negative.

Important Glossaries for the Square Root of -136

  • Imaginary Unit: The imaginary unit 'i' is defined as the square root of -1, used to express the square roots of negative numbers.
     
  • Complex Number: A number that comprises a real part and an imaginary part, typically written in the form a + bi.
     
  • Square Root: The square root of a number is a value that, when multiplied by itself, gives the original number. For negative numbers, involves the imaginary unit.
     
  • Approximation: A method of finding an estimated value close to the actual value, used here for finding square roots of non-perfect squares.
     
  • Prime Factorization: The process of expressing a number as the product of its prime factors, used in simplifying square roots.

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Jaskaran Singh Saluja

About the Author

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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