Question

The number of ordered pairs of integers (x, y) satisfying the equation ${x^2} + 6x + {y^2} = 4$ is
(a) 2
(b) 8
(c) 6
(d) None of these

Answer 1

Hint: In this question, we will first make per feet square of the possible terms by adding or subtracting to both sides of the equation. For the equation, ${x^2} + 6x + {y^2} = 4$, we can see that, we have terms ${x^2} + 6x$ and to make it perfect square, we have to add 9 on both side of the equation. Then, we will take the cases such that it satisfies the condition of the question.

Complete step-by-step solution -
We have been asked to find the number of ordered pairs of integers (x, y) which satisfies the equation ${x^2} + 6x + {y^2} = 4$.
We will complete the square for x by adding the half of coefficient of x which is 9 to both sides of the equation.
$ \Rightarrow {x^2} + 6x + {y^2} + 9 = 4 + 9$
On rearranging the terms, we get:
$ \Rightarrow {x^2} + 6x + 9 + {y^2} = 13$
Using the identity ${a^2} + 2ab + {b^2} = {\left( {a + b} \right)^2}$, we get:
$ \Rightarrow {\left( {x + 3} \right)^2} + {y^2} = 13$
Since, x and y are integers. So, ${\left( {x + 3} \right)^2}{\rm{and}}\; {y^2}$must be squares of integers less than 13, the only option is 9 and 4.
Case I: Considering ${\left( {x + 3} \right)^2} = 4$ first, we get:
$\begin{array}{l} \Rightarrow {\left( {x + 3} \right)^2} = 4\\ \Rightarrow x + 3 = \pm 2\\ \Rightarrow x = - 3 \pm 2\\ \Rightarrow x = - 5, - 1\end{array}$
Next, let us consider ${y^2} = 9$, then we will get:
\[\begin{array}{l} \Rightarrow y = \sqrt 9 \\ \Rightarrow y = \pm 3\end{array}\]
Hence, we get the ordered pair as:
$\left( { - 1,3} \right),\left( { - 5,3} \right),\left( { - 1, - 3} \right),\left( { - 5, - 3} \right)$
Case II: Considering ${\left( {x + 3} \right)^2} = 9$ first, we get:
$\begin{array}{l} \Rightarrow {\left( {x + 3} \right)^2} = 9\\ \Rightarrow x + 3 = \pm 3\\ \Rightarrow x = - 3 \pm 3\\ \Rightarrow x = - 6,0\end{array}$
Next, let us consider ${y^2} = 4$, then we will get:
\[\begin{array}{l} \Rightarrow y = \sqrt 4 \\ \Rightarrow y = \pm 2\end{array}\]
Hence, we get the ordered pair as:
$\left( {0,2} \right),\left( { - 6,2} \right),\left( {0, - 2} \right),\left( { - 6, - 2} \right)$
So, the total number of ordered pairs satisfying the equation is equal to the ordered pair in Case I and Case II.
Total number of ordered pair = 4 + 4 = 8
Therefore, the correct option is b.

Note: The first mistake that we generally do is considering the only one case out of Case I and Case II and thus, we get only half the answer. We will get only 4 ordered pairs. Then, we might choose the right option as none of these. Also, it is very important that you should remember that integers can also be negative. So, while finding the square roots, we will also consider the negative values.