Question

2304 find by factor the positive square roots.​

Answer 1

$\large\underline{\sf{Solution-}}$

Given number to find

$\rm :\longmapsto\: \sqrt{2304}$

To evaluate this, the prime factorization method is used.

So, using Prime factorization method, we have

$\begin{gathered} \:\: \begin{array}{c|c} {\underline{\sf{2}}}&{\underline{\sf{\:\:2304 \:\:}}}\\ {\underline{\sf{2}}}& \underline{\sf{\:\:1152 \:\:}} \\\underline{\sf{2}}&\underline{\sf{\:\:576\:\:}} \\ {\underline{\sf{2}}}& \underline{\sf{\:\:288 \:\:}} \\ {\underline{\sf{2}}}& \underline{\sf{\:\:144 \:\:}} \\ {\underline{\sf{2}}}& \underline{\sf{\:\:72 \:\:}} \\ {\underline{\sf{2}}}& \underline{\sf{\:\:36 \:\:}} \\ {\underline{\sf{2}}}& \underline{\sf{\:\:18 \:\:}} \\ {\underline{\sf{3}}}& \underline{\sf{\:\:9 \:\:}} \\ {\underline{\sf{3}}}& \underline{\sf{\:\:3 \:\:}}\\\underline{\sf{}}&{\sf{\:\:1 \:\:}} \end{array}\end{gathered}$

Thus,

$\rm :\longmapsto\:2304 = \underbrace{2.2}. \: \underbrace{2.2} . \: \underbrace{2.2}.\underbrace{2.2}. \: \underbrace{3.3}$

$\rm :\longmapsto\: \sqrt{2304} = \sqrt{ \underbrace{2.2}. \: \underbrace{2.2} . \: \underbrace{2.2}.\underbrace{2.2}. \: \underbrace{3.3}}$

$\rm :\longmapsto\: \sqrt{2304} = 2 \times 2 \times 2 \times 2 \times 3$

$\rm :\longmapsto\: \sqrt{2304} = 48$

Hence,

$\red{\rm \implies\:\boxed{ \tt{ \: \sqrt{2304} \: = \: 48 \: }}}$

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More to know :

1. Perfect square number can never ends with the digit 2, 3, 7, 8 and odd number of zeroes.

2. The square of even number is always even

3. The square of odd number is always odd.

4. The sum of the first 'n' odd natural number is n^2.

5. Between the squares of any two consecutive natural numbers n and (n+1), 2n natural numbers exist.

6. Number of digits in the square root of 2n digits is n.

7. Number of digits in the square root of (2n + 1) digits is (n + 1)/2.