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Is 128 a perfect number?

No, the number 128 is not a perfect square, since it cannot be expressed as a product of two equal integers.

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Square Root of 128

The square root of 128 is a number, which when multiplied by itself resulting in the original number 128. As the value of the square root of 128 cannot be expressed in the form of p/q, we can say that the square root of 128 is an irrational number. Here, we are going to discuss the two different methods such as the prime factorization method and the long division method to find the square root of 128 with complete explanation.

What is the Square Root of 128?

If a number is multiplied by itself and gives the result as 128, then the number is the square root 128. The square root of 128 is symbolically expressed as √128.

Hence, √128 = √(Number × Number)

Thus, if we multiply the number 11.3137 two times, we get the original value 128.

(i.e) √128 = √(11.3137× 11.3137)

√128 = √(11.3137)2

Now, remove square and square root, we get

√128 = ± 11.3137

Square Root of 128 in Decimal Form: 11.3137.

Square Root of 128 in Radical Form

To express the square root of 128 in the simplest radical form, write the prime factorization of 128. The prime factorization of 128 is 2×2×2×2×2×2×2. Thus, the simplest radical form of the square root of 128 is 8√2.

Square Root of 128 in Radical Form: 8√2.

Square Root of 128 by Prime Factorization Method

To find the square root of 128 using the prime factorization method, we need to know the prime factorization of 128. Thus, the prime factorization of 128 is 2×2×2×2×2×2×2.

Thus, √128 = √(2×2×2×2×2×2×2) = 8√2.

We know that,

√2 = 1.414

Now, substitute the values of √2 in the above equation.

√128 = 8(1.414)

√128 = 11.313 (approximately)

Hence, the square root of 128 in decimal form is approximately equal to 11.313.

Square Root of 128 by Long Division Method

The procedure to find the square root of 128 using the long division method is given as follows: Step 1: Write the number 128 in decimal form. To find the exact value of the square root of 128, add 6 zeros after the decimal point. Hence, 128 in decimal form is 128.000000. Now, pair the number 128 from right to left by putting the bar on the top of the number. Step 2: Now, divide the number 1 by a number, such that the product of the same number should be less than or equal to 1. Thus, 1×1 =1, which is equal to 1. Thus, we obtained the quotient = 1 and remainder = 0. Step 3: Double the quotient value, so we get 2 and assume that 20 is the new divisor. Now, bring down the value 28 for division operation. So, the new dividend obtained is 28. Now, find the number, such that (20 + new number) × new number should give the product value, that should be less than or equal to 28. Hence, (20+1) × 1 = 21, which is less than 28.

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Step 4: Now subtract 21 from 28, and we get 7 as the new reminder, and 11 as a quotient. Step 5: The new quotient obtained is 11, and double that. Hence, we get 22 and assume that 220 is our new divisor. Now, bring down the two zeros and, we have 700 as the new dividend. Step 6: Find the number, such that (220 + new number) × new number should give the product value, that should be less than or equal to 700. Thus, (220+3)× 3 = 669, which is less than 700. Step 7: Subtract 669 from 700, and we get 31 as the new reminder. Step 8: Continue this process until we get the approximate value of the square root of 128 up to three decimal places. (Note: keep the decimal point in the quotient value after bringing down all the values in the dividend). Step 9: Thus, the approximate value of the square root of 128, √128 is 11.313.

Video Lessons on Square Roots

Visualising square roots Finding Square roots

Examples

Example 1:

Simplify 8√2 ×√128.

Solution:

Given: 8√2 ×√128.

We know that the square root of 128 in radical form is 8√2.

Now, substitute the value in the expression, we get:

8√2 ×√128 = 8√2 × 8√2.

8√2 ×√128 = 64(2) = 128

Therefore, 8√2 ×√128 is 128.

Example 2:

Find the value of p, if 4p+√2√128 = 36

Solution:

Given equation: 4p+√2√128 = 36 …(1)

We know that √128 = 8√2

Now, substitute the value in equation (1),

4p+√2(8√2) = 36

4p+8(2) = 36

4p+16 = 36

4p = 36-16

4p = 20

p = 20/4 = 5

Therefore, the value of p is 5.

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