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Magazine Chapter 1: Problem 77
Solve each equation in the complex number system. $$ x^{4}+13 x^{2}+36=0 $$
Short Answer
Expert verified
The solutions are \( x = \pm 3i, \pm 2i \).
Step by step solution
01
Substitute and Simplify
Let us start by making a substitution. Let: \[ y = x^2 \]. This transforms the equation into: \[ y^2 + 13y + 36 = 0 \]
02
Factor the Quadratic Equation
Next, factor the quadratic equation: \[ y^2 + 13y + 36 = (y + 9)(y + 4) = 0 \] There are two solutions for \( y \): \[ y = -9 \] \[ y = -4 \]
03
Back-Substitute for y
Now that we have solutions for \( y \), back-substitute to solve for \( x \): \[ x^2 = -9 \] \[ x^2 = -4 \]
04
Solve for x
Solve each equation for \( x \): For \( x^2 = -9 \): \[ x = \pm 3i \] For \( x^2 = -4 \): \[ x = \pm 2i \]
05
Conclusion
The complete set of solutions for the original equation is: \[ x = \pm 3i, \pm 2i \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
complex numbers
Complex numbers are numbers that include a real part and an imaginary part. They are written in the form \(a + bi\), where \(a\) is the real part and \(b\) is the imaginary part. In mathematical operations, the imaginary unit \(i\) is defined by the property \(i^2 = -1\). Complex numbers are essential in solving equations where no real solution exists. In this exercise, we found solutions like \(x = 3i\) and \(x = 2i\), which are pure imaginary numbers, indicating they lie on the imaginary axis of the complex plane.
Here are some key points about complex numbers:
Here are some key points about complex numbers:
- The real part (\(a\)) can be zero.
- The imaginary part (\(bi\)) can also be zero, in which case the complex number is actually a real number.
- If both the real and imaginary parts are non-zero, the number is truly complex.
quadratic equations
A quadratic equation is any equation that can be rearranged in standard form as \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants. The equation in our example was transformed into a quadratic by substituting \(y = x^2\).
This substitution simplified our problem into a form more easily handled using known methods for solving quadratic equations.
To solve quadratic equations:
This substitution simplified our problem into a form more easily handled using known methods for solving quadratic equations.
To solve quadratic equations:
- You can use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
- Sometimes, it's possible to factor the quadratic equation, which is what we did in this problem (\((y + 9)(y + 4) = 0\)).
factoring
Factoring is the process of breaking down an expression into simpler multipliers that produce the original expression when multiplied together. In our exercise, the transformed quadratic equation \(y^2 + 13y + 36\) was factored into \((y + 9)(y + 4)\).
This was crucial because it allowed us to find the roots of the equation.
Here's the process:
This was crucial because it allowed us to find the roots of the equation.
Here's the process:
- Write the quadratic equation in standard form: \(y^2 + 13y + 36 = 0\).
- Look for two numbers that multiply to give you the constant term (36) and add to give you the linear coefficient (13). These numbers are 9 and 4.
- Factor the equation as \((y + 9)(y + 4) = 0\).
- Solve each factor for y: \(y + 9 = 0\) and \(y + 4 = 0\).
- This gives the solutions \(y = -9\) and \(y = -4\).
substitution method
The substitution method is a technique used to simplify complex equations by replacing variables with simpler expressions. In our exercise, we used substitution to turn a quartic equation into a quadratic one.
We replaced \(x^2\) with \(y\), reducing our original equation \(x^4 + 13x^2 + 36 = 0\) to \(y^2 + 13y + 36 = 0\).
Steps for substitution:
We replaced \(x^2\) with \(y\), reducing our original equation \(x^4 + 13x^2 + 36 = 0\) to \(y^2 + 13y + 36 = 0\).
Steps for substitution:
- Identify a substitution that simplifies the equation. For example, let \(y = x^2\), which reduces the power of our variable.
- Rewrite the equation in terms of the new variable.
- Solve the simpler equation using appropriate methods (factoring, quadratic formula).
- Back-substitute the original variable to find the solutions to the original problem.
