Systems of Quadratic Equations in Two Variables
An equation of the form
\[
ax^2 + bxy + cy^2 + dx + ey + f = 0
\]
where \( a, b, c, d, e, f \in \mathbb{R} \) and at least one of the coefficients \( a, b, \) or \( c \) is non-zero, is called a quadratic equation in two variables.
A set containing at least two equations where at least one equation is quadratic in two variables is called a system of quadratic equations in two variables.
The set of all ordered pairs of real numbers \( (x, y) \) that satisfy all equations simultaneously is called the solution set of the system.
To solve these systems of equations, we typically use substitution or elimination to obtain a single-variable equation from the given equations.
Example:
Find the solution set of the following system of equations:
\[
x^2 + xy + y^2 = 21
\]
\[
x + y = 3\sqrt{3}
\]
Isolating \( y \) in the second equation gives:
\[ x + y = 3\sqrt{3} \Rightarrow y = 3\sqrt{3} – x \]
Substituting this expression into the first equation:
\[
x^2 + xy + y^2 = 21
\]
\[
\Rightarrow x^2 + x (3\sqrt{3} – x) + (3\sqrt{3} – x)^2 = 21
\]
\[
\Rightarrow x^2 – 3\sqrt{3}x + 6 = 0
\]
Factoring the quadratic equation:
\[
\Rightarrow (x – \sqrt{3}) (x – 2\sqrt{3}) = 0
\]
\[
\Rightarrow x_1 = \sqrt{3} \quad \text{or} \quad x_2 = 2\sqrt{3}
\]
Now, we substitute these values back into the linear equation to find the corresponding \( y \)-values.
For \( x_1 = \sqrt{3} \) in \( x + y = 3\sqrt{3} \):
\[
\sqrt{3} + y = 3\sqrt{3} \Rightarrow y_1 = 2\sqrt{3}
\]
For \( x_2 = 2\sqrt{3} \):
\[
2\sqrt{3} + y = 3\sqrt{3} \Rightarrow y_2 = \sqrt{3}
\]
Thus, the solution set is:
\[
S = \{ (\sqrt{3}, 2\sqrt{3}), (2\sqrt{3}, \sqrt{3}) \}
\]
Example:
Find the solution set of the following system of equations:
\[
3x^2 + 2y^2 = 14
\]
\[
x^2 + y^2 = 5
\]
Multiply both sides of the second equation by -2 and eliminate \( y^2 \) by adding the equations:
\[
3x^2 + 2y^2 = 14
\]
\[
+ (-2x^2 – 2y^2 = -10)
\]
\[
x^2 = 4 \Rightarrow x_1 = -2 \quad \text{or} \quad x_2 = 2
\]
Substituting these values back into the second equation:
In \( x^2 + y^2 = 5 \), for \( x_1 = -2 \):
\[ 4 + y^2 = 5 \]
\[
\Rightarrow y_1 = -1 \quad \text{or} \quad y_2 = 1
\]
For \( x_2 = 2 \):
\[ 4 + y^2 = 5 \]
\[
\Rightarrow y_3 = -1 \quad \text{or} \quad y_4 = 1
\]
Therefore, the solution set is:
\[
S = \{ (-2, -1), (-2, 1), (2, -1), (2, 1) \}
\]
Example:
Find the solution set of the following system of equations:
\[
3x^2 + 2y^2 = 14
\]
\[
x^2 + y^2 = 5
\]
Multiply both sides of the second equation by -2:
\[\begin{aligned}
3x^2 + 2y^2 = 14 \\
+ \quad \quad -2x^2 – 2y^2 = -10 \\
\hline \\
x^2 = 4 \\
\Rightarrow x_1 = -2 \quad \text{or} \quad x_2 = 2
\end{aligned}
\]
Substituting these values back into the second equation:
In \( x^2 + y^2 = 5 \), for \( \quad x_1 = -2 \):
\[ \quad 4 + y^2 = 5 \]
\[
\Rightarrow y_1 = -1 \quad \text{or} \quad y_2 = 1
\]
For \( x_2 = 2 \):
\[ \quad 4 + y^2 = 5 \]
\[
\Rightarrow y_3 = -1 \quad \text{or} \quad y_4 = 1
\]
Thus, the solution set is:
\[
S = \{ (-2, -1), (-2, 1), (2, -1), (2, 1) \}
\]
Example:
Find the solution set of the following system of equations:
\[
|x + 2y| = 1
\]
\[
x + y = 2
\]
Isolating \( x \) in the second equation gives \( x = 2 – y \). Substituting this into the first equation:
\[
|x + 2y| = 1 \Rightarrow |2 – y + 2y| = 1
\]
Squaring both sides of the absolute value equation:
\[
\Rightarrow |2 + y|^2 = 1^2
\]
\[
\Rightarrow 4 + 4y + y^2 = 1
\]
\[
\Rightarrow y^2 + 4y + 3 = 0
\]
Solving the quadratic equation:
\[
\Rightarrow y = -3 \quad \text{or} \quad y = -1
\]
Substitute these values back into the linear equation \( x + y = 2 \):
For \( y = -3 \):
\[
x – 3 = 2 \Rightarrow x = 5
\]
For \( y = -1 \):
\[
x – 1 = 2 \Rightarrow x = 3
\]
Thus, the solution set is:
\[
S = \{ (5, -3), (3, -1) \}
\]
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