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Chapter 5: Problem 393

The number of ordered pairs of integers \((x, y)\) satisfying the equation \(x^{2}+6 x+y^{2}=4\) is (a) 2 (b) 4 (c) 6 (d) 8

Short Answer

Expert verified
The number of ordered pairs of integers (x, y) satisfying the equation \(x^2 +6x + y^2 = 4\) is 4. The solutions are (0, 2), (0, -2), (-6, 2), and (-6, -2).

Step by step solution

01

Rearrange the equation

In order to get better insight into the possible solutions, we can rearrange the equation as follows: \(x^2 + 6x + y^2 = 4\) \((x^2 + 6x + 9) + y^2 = 13\) \((x + 3)^2 + y^2 = 13\)
02

Visualize the equation

The rearranged equation in Step 1 resembles the equation of a circle in a Cartesian plane given by \((x-a)^2 + (y-b)^2 = r^2\). In our case, we have a circle centered at (-3, 0) with radius sqrt(13). Now we need to find the lattice points (integer coordinates) that lie on this circle.
03

Test integer values for x and y

Since the equation is symmetric with respect to both x and y, we can only look for solutions in one quadrant and then derive the corresponding solutions in the other quadrants. Let's check Quadrant I (where both x and y are positive or zero): Based on the equation \((x+3)^2 + y^2 = 13\) : - If x = 0, then (-3)^2 + y^2 = 13 => y^2 = 4 => y = ±2 - If y = 0, then (x+3)^2 = 13 => x = ±√(10) - 3, which is not an integer. We conclude that (0,2) and (0,-2) are solutions. Now, we can easily find the other symmetric solutions: - Reflection across y-axis: (-6,2) and (-6,-2) - Reflection across x-axis: (0,-2) and (-6,2) - Reflection across both axes: (0,2) and (-6,-2)
04

Count the solutions

We have found a total of 4 unique ordered pairs of integers (x, y) that satisfy the given equation: (0,2), (0,-2), (-6,2), and (-6,-2). Therefore, the correct answer is: (b) 4

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