Question

The least integer that must be added to $\left( {9798 \times 9792} \right)$ to make it a perfect square is:
A) 9
B) 8
C) 7
D) 6

Answer 1

Hint:
We will write 9798 in terms of 9792 and then, we will make it a perfect square in terms of 9792. We will simplify the terms in the form of the algebraic identity ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$ to make it a perfect square.

Complete step by step solution:
We are given that we need to find a least integer value that must be added to make $\left( {9798 \times 9792} \right)$ a perfect square.
Now, we can write 9798 in terms of 9792 as: $9798 = 9792 + 6$
$ \Rightarrow $ $\left( {9798 \times 9792} \right) = (9792 + 6) \times 9792$
We can write this equation after opening the bracket as:
$ \Rightarrow \left( {9798 \times 9792} \right)= {\left( {9792} \right)^2}+ 6 \times 9792$
We can split 6 as $2 \times 3$. The equation becomes
$ \Rightarrow {\left( {9792} \right)^2}+ 6 \times 9792 = {\left( {9792} \right)^2}+ 2 \times 3 \times 9792$
Comparing it with the algebraic identity: ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$, we can say that $a = 9792, b = 3$. We only need $b^2$ to make this equation a perfect square, i.e., we need $(3)^{2} = 9$.
$ \Rightarrow {\left( {9792} \right)^2}+ 6 \times 9792 = {\left( {9792} \right)^2}+ 2 \times 3 \times 9792 + (3)^{2}$

Therefore, we need to add 9 to make $\left( {9798 \times 9792} \right)$ a perfect square.
Hence, option (A) is correct.

Note:
In this question, you may go confused in the step while splitting 9798 in terms of 9792 in order to reduce it in a term closer to make it a perfect square. You may go wrong while using the algebraic identity ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$ to make the term ${\left( {9792} \right)^2}+ 2 \times 3 \times 9792$ a perfect square.