Problem 70 - Precalculus Mathematics for Calculus

Question

So far, we have worked only with polynomials that have real coefficients. These exercises involve polynomials with real and imaginary coefficients. Find all solutions of the equation. (a) \(2 x+4 i=1\) (b) \(x^{2}-i x=0\) (c) \(x^{2}+2 i x-1=0\) (d) \(i x^{2}-2 x+i=0\)

Step-by-Step Solution

Answer

(a) \(x = \frac{1}{2} - 2i\), (b) \(x = 0\), \(x = i\), (c) \(x = -i\), (d) \(x = -i \pm i\sqrt{2}\).

Step 1: Solve Equation (a)

We start with the equation \(2x + 4i = 1\). Rearrange the equation to isolate \(x\): \[2x = 1 - 4i\] Divide both sides by 2: \[x = \frac{1}{2} - 2i\] Thus, the solution to equation (a) is \(x = \frac{1}{2} - 2i\).

Step 2: Solve Equation (b)

The equation is \(x^2 - ix = 0\). Factor the left side: \[x(x - i) = 0\] Based on the zero product property, either \(x = 0\) or \(x - i = 0\). Solve for \(x\): - \(x = 0\) - \(x = i\) Therefore, the solutions are \(x = 0\) and \(x = i\).

Step 3: Solve Equation (c)

The equation is \(x^2 + 2ix - 1 = 0\). This is a quadratic equation, so use the quadratic formula: \[x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a}\] Here, \(a = 1\), \(b = 2i\), and \(c = -1\). Substitute these values into the formula: \[x = \frac{{-2i \pm \sqrt{{(2i)^2 - 4(1)(-1)}}}}{2}\] Calculate under the square root: \[(2i)^2 = -4\] \[-4(-1) = 4\] \[b^2 - 4ac = -4 + 4 = 0\] So, \[x = \frac{{-2i \pm \sqrt{0}}}{2}\] which simplifies to: \[x = -i\] Thus, the solution is \(x = -i\).

Step 4: Solve Equation (d)

Given the equation \(ix^2 - 2x + i = 0\), we utilize the quadratic formula: \[x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a}\] In this equation, \(a = i\), \(b = -2\), and \(c = i\). Substitute in these values: \[x = \frac{{2 \pm \sqrt{{(-2)^2 - 4(i)(i)}}}}{2i}\] Calculate under the square root: \[(-2)^2 = 4\] \[4i^2 = -4\] \[b^2 - 4ac = 4 - (-4) = 8\] \[\sqrt{8} = 2\sqrt{2}\] Now substitute back: \[x = \frac{{2 \pm 2\sqrt{2}}}{2i}\] Which simplifies to \[x = \frac{1 \pm \sqrt{2}}{i}\] To simplify \(\frac{1}{i}\), multiply numerator and denominator by \(-i\): \[x = -i(1 \pm \sqrt{2})\] Therefore, the solutions are \(x = -i \pm i\sqrt{2}\).

Key Concepts

Imaginary Coefficients

Imaginary coefficients are found in polynomials when the numbers are multiplied by the imaginary unit, denoted as \( i \), where \( i^2 = -1 \).
This imaginary unit allows for the representation of numbers that cannot be denoted on the real number line, such as square roots of negative numbers.
In polynomial equations, coefficients can be either real numbers, like 2 or 3, or imaginary, like \( 4i \) or \( -i \).
This adds a layer of complexity and opens up the world of complex numbers, where these equations exist.

• Example: In equation (a) from the exercise, the polynomial \(2x + 4i = 1\) features the imaginary coefficient \(4i\).
• Imaginary coefficients often require students to perform operations involving the imaginary unit \(i\), which is crucial for solving complex polynomial equations effectively.

Understanding and working with imaginary coefficients is key when dealing with many advanced mathematical concepts.

Quadratic Formula

The quadratic formula is a powerful tool for solving quadratic equations, which are equations of the form \(ax^2 + bx + c = 0\).
It provides a way to find the values of \(x\) that satisfy this equation, using the formula:\[x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a}\]Here, \(a\), \(b\), and \(c\) are coefficients from the polynomial, which can be real or imaginary numbers.

• Example: In our exercise, equation (c), \(x^2 + 2ix - 1 = 0\), employs the quadratic formula to find the solution \(x = -i\).
• Even when coefficients are imaginary, the quadratic formula can still be used to find the solutions by including \(i\) in calculations, particularly inside the discriminant \(b^2 - 4ac\).

Using the quadratic formula can simplify the process of solving these equations compared to factoring or other methods, especially when dealing with more complex or non-factorable polynomials.

Zero Product Property

The zero product property is a fundamental principle in algebra that states if a product of two numbers equals zero, then at least one of the multipliers must also be zero.
It is particularly useful for solving equations that are expressed as products, typically after factoring a polynomial.

• Example: Consider equation (b) from the exercise: \(x^2 - ix = 0\). This can be factored into \(x(x - i) = 0\).
• Using the zero product property, the solutions are \(x = 0\) and \(x = i\), as these are the values that make each factor zero.

This property simplifies solving equations by turning a potentially complex equation into simpler, linear factors that can be easily solved.

Polynomials

Polynomials are algebraic expressions consisting of variables and coefficients, where the variables are raised to whole number exponents.
They form the basis for many algebraic operations and can range from simple linear expressions to complex equations involving imaginary coefficients.
Polynomials can have various degrees, which are defined by the highest exponent of the variable in the expression.

• Characterization: The polynomial \(x^2 - ix = 0\) from the exercise is a quadratic polynomial because its highest degree is 2.
• Diverse Coefficients: In exercises like those provided, coefficients can involve imaginary numbers, such as in \(x^2 + 2ix - 1 = 0\).

Polynomials are a crucial part of mathematics because they are used to model various real-life problems and allow for the exploration of more abstract mathematical ideas.