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Magazine Chapter 9: Problem 12
Give all solutions of each nonlinear system of equations, including those with nonreal complex components. $$\begin{aligned} &y=x^{2}+6 x+9\\\ &x+2 y=-2 \end{aligned}$$
Short Answer
Expert verified
The solutions are (-2.5, 0.25) and (-4, 1).
Step by step solution
01
- Identify the equations in the system
The system is composed of two equations:1. \( y = x^2 + 6x + 9 \)2. \( x + 2y = -2 \)
02
- Substitute the expression for y into the second equation
Substitute \( y = x^2 + 6x + 9 \) into the second equation \( x + 2y = -2 \):\[ x + 2(x^2 + 6x + 9) = -2 \]
03
- Simplify the substituted equation
Simplify the equation obtained in Step 2:\[ x + 2x^2 + 12x + 18 = -2 \]Combine like terms:\[ 2x^2 + 13x + 18 = -2 \]
04
- Move all terms to one side to set the equation to zero
Move -2 to the left side:\[ 2x^2 + 13x + 18 + 2 = 0 \]\[ 2x^2 + 13x + 20 = 0 \]
05
- Solve the quadratic equation
Solve the quadratic equation \( 2x^2 + 13x + 20 = 0 \) using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 2 \), \( b = 13 \), and \( c = 20 \):\[ x = \frac{-13 \pm \sqrt{13^2 - 4(2)(20)}}{2(2)} \]\[ x = \frac{-13 \pm \sqrt{169 - 160}}{4} \]\[ x = \frac{-13 \pm 3}{4} \]This gives two solutions:\[ x = \frac{-13 + 3}{4} = -2.5 \]\[ x = \frac{-13 - 3}{4} = -4 \]
06
- Find the corresponding y values
Substitute the x values back into the first equation to get y:For \( x = -2.5 \):\[ y = (-2.5)^2 + 6(-2.5) + 9 \]\[ y = 6.25 - 15 + 9 \]\[ y = 0.25 \]For \( x = -4 \):\[ y = (-4)^2 + 6(-4) + 9 \]\[ y = 16 - 24 + 9 \]\[ y = 1 \]
07
- Compile the solutions
The solutions to the system are the pairs:\( (-2.5, 0.25) \) and \( (-4, 1) \)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
substitution method
The substitution method is a crucial technique for solving system of equations. It involves solving one equation for one variable and then substituting that expression into another equation. This simplifies the system into a single equation with one variable.
Example: In the given system, we first solve the equation for \(y\):
\(y = x^2 + 6x + 9\).
This relation is then substituted into the second equation, replacing y:
\(x + 2y = -2\)
\(x + 2(x^2 + 6x + 9) = -2\).
Now, we have an equation with just one variable, \(x\). This allows us to further simplify and solve it.
Example: In the given system, we first solve the equation for \(y\):
\(y = x^2 + 6x + 9\).
This relation is then substituted into the second equation, replacing y:
\(x + 2y = -2\)
\(x + 2(x^2 + 6x + 9) = -2\).
Now, we have an equation with just one variable, \(x\). This allows us to further simplify and solve it.
quadratic formula
The quadratic formula is a standard method for finding the solutions to a quadratic equation: \(ax^2 + bx + c = 0\). The formula is:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \]
where \(a\), \(b\), and \(c\) are coefficients in the quadratic equation.
Example: In our problem, after substitution and simplification, we obtain: \(2x^2 + 13x + 20 = 0\).
Here, \(a = 2\), \(b = 13\), and \(c = 20\). Using the quadratic formula, we calculate:
\[ x = \frac{-13 \pm \sqrt{169 - 160}}{4}, \]
which results in two solutions for \(x\): \(x = -2.5\) and \(x = -4\).
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \]
where \(a\), \(b\), and \(c\) are coefficients in the quadratic equation.
Example: In our problem, after substitution and simplification, we obtain: \(2x^2 + 13x + 20 = 0\).
Here, \(a = 2\), \(b = 13\), and \(c = 20\). Using the quadratic formula, we calculate:
\[ x = \frac{-13 \pm \sqrt{169 - 160}}{4}, \]
which results in two solutions for \(x\): \(x = -2.5\) and \(x = -4\).
complex solutions
Complex solutions occur when the discriminant (the part under the square root in the quadratic formula: \(b^2 - 4ac\) ) is negative. This results in the square root of a negative number, which introduces imaginary numbers.
Although this particular problem does not present complex solutions, it is essential to be prepared for them. When the discriminant is negative, we rely on imaginary numbers, denoted usually by 'i', where \(i^2 = -1\). For example, if \(b^2 - 4ac\) were negative, say \(-9\), we would write: \(\sqrt{-9} = 3i\).
Understanding how to handle complex solutions is vital, as they frequently appear in advanced problems.
Although this particular problem does not present complex solutions, it is essential to be prepared for them. When the discriminant is negative, we rely on imaginary numbers, denoted usually by 'i', where \(i^2 = -1\). For example, if \(b^2 - 4ac\) were negative, say \(-9\), we would write: \(\sqrt{-9} = 3i\).
Understanding how to handle complex solutions is vital, as they frequently appear in advanced problems.
identifying equations
To solve any system of equations, it's essential first to correctly identify each equation and understand the relationships between variables.
Here, the system comprises:
By identifying each equation and its structure, we grasp how they interact.
Example: The first equation is a quadratic equation in terms of \(x\) and defines \(y\) in terms of \(x\). The second equation is linear, relating \(x\) to \(2y\). This clarity aids in deciding the right solving strategy.
Here, the system comprises:
- First equation: \(y = x^2 + 6x + 9\)
- Second equation: \(x + 2y = -2\)
By identifying each equation and its structure, we grasp how they interact.
Example: The first equation is a quadratic equation in terms of \(x\) and defines \(y\) in terms of \(x\). The second equation is linear, relating \(x\) to \(2y\). This clarity aids in deciding the right solving strategy.
simplifying equations
Simplifying equations is about combining like terms and making the equation easier to manage.
In our example, after substituting \(y\) into the second equation, we simplify:
\(x + 2x^2 + 12x + 18 = -2\)
By combining like terms, it becomes:
\(2x^2 + 13x + 18 = -2\).
Moving all terms to one side sets the equation to zero:
\(2x^2 + 13x + 20 = 0\).
Simplification condenses our problem into a solvable form, reducing potential mistakes and making the solving process clearer.
In our example, after substituting \(y\) into the second equation, we simplify:
\(x + 2x^2 + 12x + 18 = -2\)
By combining like terms, it becomes:
\(2x^2 + 13x + 18 = -2\).
Moving all terms to one side sets the equation to zero:
\(2x^2 + 13x + 20 = 0\).
Simplification condenses our problem into a solvable form, reducing potential mistakes and making the solving process clearer.
