The square root of $1764$ is
$A)42$
$B)82$
$C)76$
$D)14$
Answer
527.4k+ views
Hint: First, we form the indicated formula for the value square root $2$ that is a square root of $1764$. And then we will find its prime factorization of the number $1764$. Then we take the one digit out of the two same number of primes. Prime numbers are the numbers that are divisible by themselves and $1$ only or also known as the numbers whose factors are the given number itself.
But the composite numbers which are divisible by themselves, $1$ and also with some other numbers (at least one number other than $1$ and itself)
Every composite number can be represented in the form of the prime factorization.
Complete step by step answer:
Since given that square root of $1764$ is represented mathematically as $\sqrt[2]{{1764}}$
Now by the prime factorization method we have $1764$ is a composite number and which can be generalized as
$
2\left| \!{\underline {\,
{1764} \,}} \right. \\
2\left| \!{\underline {\,
{882} \,}} \right. \\
3\left| \!{\underline {\,
{441} \,}} \right. \\
3\left| \!{\underline {\,
{147} \,}} \right. \\
7\left| \!{\underline {\,
{49} \,}} \right. \\
7\left| \!{\underline {\,
7 \,}} \right. \\
1\left| \!{\underline {\,
1 \,}} \right. \\
$
Thus, we have $2 \times 2 \times 3 \times 3 \times 7 \times 7$ and by multiplication we get $2 \times 2 \times 3 \times 3 \times 7 \times 7 = 1764$
Hence the square root of $1764$ is $\sqrt[2]{{1764}} = \sqrt[2]{{2 \times 2 \times 3 \times 3 \times 7 \times 7}}$
Now $\sqrt[2]{{a \times a}} = a$ hence we get $\sqrt[2]{{1764}} = \sqrt[2]{{2 \times 2 \times 3 \times 3 \times 7 \times 7}} \Rightarrow 2 \times 3 \times 7$
Again, by the multiplication we get $\sqrt[2]{{1764}} = 42$
So, the correct answer is “Option A”.
Note:
Since don’t write the number ${3^2} = 9$ because $9$ is the composite number, and thus ${3^2}$ is the repeated prime number while in the process of prime factorization.
We can find if a given number is prime or composite by the trial-and-error methods. Divide the number with any prime numbers less than the given number. if the number is divisible by the prime number, it is the composite number, if not then it is the prime number. The only even prime number is $2$ and all other prime numbers are odd.
But the composite numbers which are divisible by themselves, $1$ and also with some other numbers (at least one number other than $1$ and itself)
Every composite number can be represented in the form of the prime factorization.
Complete step by step answer:
Since given that square root of $1764$ is represented mathematically as $\sqrt[2]{{1764}}$
Now by the prime factorization method we have $1764$ is a composite number and which can be generalized as
$
2\left| \!{\underline {\,
{1764} \,}} \right. \\
2\left| \!{\underline {\,
{882} \,}} \right. \\
3\left| \!{\underline {\,
{441} \,}} \right. \\
3\left| \!{\underline {\,
{147} \,}} \right. \\
7\left| \!{\underline {\,
{49} \,}} \right. \\
7\left| \!{\underline {\,
7 \,}} \right. \\
1\left| \!{\underline {\,
1 \,}} \right. \\
$
Thus, we have $2 \times 2 \times 3 \times 3 \times 7 \times 7$ and by multiplication we get $2 \times 2 \times 3 \times 3 \times 7 \times 7 = 1764$
Hence the square root of $1764$ is $\sqrt[2]{{1764}} = \sqrt[2]{{2 \times 2 \times 3 \times 3 \times 7 \times 7}}$
Now $\sqrt[2]{{a \times a}} = a$ hence we get $\sqrt[2]{{1764}} = \sqrt[2]{{2 \times 2 \times 3 \times 3 \times 7 \times 7}} \Rightarrow 2 \times 3 \times 7$
Again, by the multiplication we get $\sqrt[2]{{1764}} = 42$
So, the correct answer is “Option A”.
Note:
Since don’t write the number ${3^2} = 9$ because $9$ is the composite number, and thus ${3^2}$ is the repeated prime number while in the process of prime factorization.
We can find if a given number is prime or composite by the trial-and-error methods. Divide the number with any prime numbers less than the given number. if the number is divisible by the prime number, it is the composite number, if not then it is the prime number. The only even prime number is $2$ and all other prime numbers are odd.
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