Question

Show that each of the following numbers is a perfect square. In each case, find the whose square number is the given
(i)1225 (ii)2601 (iii)5929 (iv)7056 (v)8281

Answer 1

Hint: A square root of a number is a value that can be multiplied by itself to give the original number. It is a function of $y = \sqrt x $. The principal square root is the positive number square root. Unless “the square root of a number refers only to the principal square root. The square root is usually denoted as $'\sqrt {} '$ and it is known as a radical symbol.

Complete step-by-step solution:
Given,
(i) 1225
First of all we have to factorize the number $1225$
$ \Rightarrow 1225 = 5 \times 5 \times 7 \times 7$
Now we have to make the set and take out the radical.
The square number is, 1225.
$ = \sqrt {1225} $
$ = \sqrt {5 \times 5 \times 7 \times 7} $
$ = \sqrt {35 \times 35} $
Now make the set and take out from the radical.
$ = 35$
Hence $1225$ is the perfect square of $35.$
(ii) 2601
First of all we have to factorize the number $2601$
$ \Rightarrow 2601 = 3 \times 3 \times 17 \times 17$
Now we have to make the set and take out the radical.
The square number is, $2601$.
$ = \sqrt {2601} $
$ = \sqrt {3 \times 3 \times 17 \times 17} $
$ = \sqrt {51 \times 51} $
Now make the set and take out from the radical.
$ = 51$
Hence $2601$ is the perfect square of $51.$
(iii) 5929
First of all we have to factorize the number $5929$
$ \Rightarrow 5929 = 7 \times 7 \times 11 \times 11$
Now we have to make the set and take out the radical.
The square number is, $5929$
$ = \sqrt {5929} $
$ = \sqrt {7 \times 7 \times 11 \times 11} $
$ = \sqrt {77 \times 77} $
Now make the set and take out from the radical.
$ = 77$
Hence $5929$ is the perfect square of $77.$
(iv) 7056
First of all we have to factorize the number $7056$
$ \Rightarrow 7056 = 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 7 \times 7$
Now we have to make the set and take out the radical.
The square number is, $7056$
$ = \sqrt {7056} $
$ = \sqrt {2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 7 \times 7} $
$ = \sqrt {84 \times 84} $
Now make the set and take out from the radical.
$ = 84$
Hence $7056$ is the perfect square of $84.$
(v) 8281
First of all we have to factorize the number $8281$
$ \Rightarrow 8281 = 13 \times 13 \times 7 \times 7$
Now we have to make the set and take out the radical.
The square number is, $8281$
$ = \sqrt {8281} $
$ = \sqrt {13 \times 13 \times 7 \times 7} $
$ = \sqrt {91 \times 91} $
Now make the set and take out from the radical.
$ = 91$
Hence $8281$ is the perfect square of $91.$

Note: The square root property is the one method to solve the quadratic equations. The method is generally used in equations that have form $a{x^2} = c$ or ${(ax + b)^2} = c$ the equation which can be expressed on either of those forms. We can take square roots from both sides. The square root of a negative number is known as a complex number.