
Find the square root of 64.
Answer
610.2k+ views
Hint: Find the prime factorisation of 64. Use the fact that if the prime factorisation of a number is ${{p}_{1}}^{{{\alpha }_{1}}}\times {{p}_{2}}^{{{\alpha }_{2}}}\times \cdots \times {{p}_{n}}^{{{\alpha }_{n}}}$ then the square root of the number is given by ${{p}_{1}}^{\dfrac{{{\alpha }_{1}}}{2}}\times {{p}_{2}}^{\dfrac{{{\alpha }_{2}}}{2}}\times \cdots \times {{p}_{n}}^{\dfrac{{{\alpha }_{n}}}{2}}$. Hence find the square root of the number.
Complete step-by-step answer:
Before solving the above problem, we need to understand what the square root of the number means.
Consider the equation ${{x}^{2}}=a,a > 0$
Clearly, there exist two values of x for which the equation is satisfied. These two values are called square roots of a. The positive root among the two roots is called principal square root, or simply the square root of a denoted as $\sqrt{a}$.
Finding the square root of a number:
When we want to find the square root of a number, we first find the prime factorisation of the number. Then we make use of the identity ${{\left( ab \right)}^{n}}={{a}^{n}}{{b}^{n}}$ and hence find the square root of the number. Since according to the unique factorisation theorem, prime factorisation of a number is unique, we have the square root of the number is also unique.
Finding the square root of 64:
We have $64={{2}^{6}}$, which is the required prime factorisation of 64.
Now we know that if the prime factorisation of a number is ${{p}_{1}}^{{{\alpha }_{1}}}\times {{p}_{2}}^{{{\alpha }_{2}}}\times \cdots \times {{p}_{n}}^{{{\alpha }_{n}}}$ then the square root of the number is given by ${{p}_{1}}^{\dfrac{{{\alpha }_{1}}}{2}}\times {{p}_{2}}^{\dfrac{{{\alpha }_{2}}}{2}}\times \cdots \times {{p}_{n}}^{\dfrac{{{\alpha }_{n}}}{2}}$.
Hence, we have $\sqrt{64}={{2}^{\dfrac{6}{2}}}={{2}^{3}}=8$
Hence the square root of 64 is 8.
Note: Verification:
Since 8 is the square root of the number, it must satisfy two properties
[i] 8 must be non-negative, which is clearly true
[ii] 8 must satisfy ${{x}^{2}}=64$. Since ${{8}^{2}}=64$, 8 satisfies ${{x}^{2}}=64$. Hence the square root of 64 is 8.
Complete step-by-step answer:
Before solving the above problem, we need to understand what the square root of the number means.
Consider the equation ${{x}^{2}}=a,a > 0$
Clearly, there exist two values of x for which the equation is satisfied. These two values are called square roots of a. The positive root among the two roots is called principal square root, or simply the square root of a denoted as $\sqrt{a}$.
Finding the square root of a number:
When we want to find the square root of a number, we first find the prime factorisation of the number. Then we make use of the identity ${{\left( ab \right)}^{n}}={{a}^{n}}{{b}^{n}}$ and hence find the square root of the number. Since according to the unique factorisation theorem, prime factorisation of a number is unique, we have the square root of the number is also unique.
Finding the square root of 64:
We have $64={{2}^{6}}$, which is the required prime factorisation of 64.
Now we know that if the prime factorisation of a number is ${{p}_{1}}^{{{\alpha }_{1}}}\times {{p}_{2}}^{{{\alpha }_{2}}}\times \cdots \times {{p}_{n}}^{{{\alpha }_{n}}}$ then the square root of the number is given by ${{p}_{1}}^{\dfrac{{{\alpha }_{1}}}{2}}\times {{p}_{2}}^{\dfrac{{{\alpha }_{2}}}{2}}\times \cdots \times {{p}_{n}}^{\dfrac{{{\alpha }_{n}}}{2}}$.
Hence, we have $\sqrt{64}={{2}^{\dfrac{6}{2}}}={{2}^{3}}=8$
Hence the square root of 64 is 8.
Note: Verification:
Since 8 is the square root of the number, it must satisfy two properties
[i] 8 must be non-negative, which is clearly true
[ii] 8 must satisfy ${{x}^{2}}=64$. Since ${{8}^{2}}=64$, 8 satisfies ${{x}^{2}}=64$. Hence the square root of 64 is 8.
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