Problem 1 - Algebra 2 and Trigonometry

Question

Emily said that when \(a\) and \(c\) are real numbers with the same sign and \(b=0\) , the roots of the equation \(a x^{2}+b x+c=0\) are pure imaginary. Do you agree with Emily? Justify your answer.

Step-by-Step Solution

Answer

Yes, Emily is correct; the roots are pure imaginary.

Step 1: Identify the Quadratic Equation

The general form of a quadratic equation is \(ax^2 + bx + c = 0\). Given the condition \(b = 0\), the equation simplifies to \(ax^2 + c = 0\).

Step 2: Isolate the Quadratic Term

Rearrange the equation to isolate the term involving \(x^2\): \(ax^2 = -c\).

Step 3: Solve for \(x^2\)

Divide both sides by \(a\) (assuming \(aeq0\)) to get \(x^2 = -\frac{c}{a}\).

Step 4: Analyze the Sign of \(-\frac{c}{a}\)

Since \(a\) and \(c\) have the same sign, their ratio \(\frac{c}{a}\) is positive. Therefore, \(-\frac{c}{a}\) is negative.

Step 5: Find the Roots

The solutions for \(x\) are given by \(x = \pm \sqrt{-\frac{c}{a}}\). Since \(-\frac{c}{a}\) is negative, \(\sqrt{-\frac{c}{a}}\) involves the square root of a negative number, leading to imaginary roots.

Key Concepts

Quadratic Equation

A quadratic equation is a fundamental concept in algebra that describes a polynomial equation of degree two. In its general form, it is written as:\[ ax^2 + bx + c = 0 \]where:

• \(a\), \(b\), and \(c\) are constants with \(a eq 0\).
• \(x\) represents the variable or unknown we need to solve for.

The term \(ax^2\) is called the quadratic term, \(bx\) is the linear term, and \(c\) is the constant term. Quadratic equations can be solved using a variety of methods, such as factoring, completing the square, or using the quadratic formula.In this exercise's scenario, we particularly focus on situations where \(b = 0\), simplifying our equation to \(ax^2 + c = 0\). This is a simpler form and provides interesting insights into the nature of the roots, especially when discussing imaginary solutions.

Real Numbers

Real numbers consist of all the numbers that can be found on the number line. This includes both rational numbers (such as fractions and integers) and irrational numbers (such as \(\sqrt{2}\) or \(\pi\)). Essentially, real numbers encompass:

• Whole numbers: like 0, 1, 2, ...
• Negative numbers: like -1, -2, ...
• Fractions: such as \(\frac{1}{2}\), \(\frac{3}{4}\)
• Irrational numbers: numbers that cannot be expressed as a fraction of two integers

In the context of our problem, when \(a\) and \(c\) are stated to be real numbers with the same sign, it implies that we are dealing with coefficients that can be placed on a real number line and verge into imaginary solutions. The real number properties are crucial in determining the nature of the quadratic equation’s roots.

Pure Imaginary

When we talk about pure imaginary numbers, they are numbers that involve the imaginary unit \(i\), where \(i\) is defined as the square root of -1:\\[ i^2 = -1 \]A number is considered pure imaginary if it is a real number multiplied by \(i\), such as \(3i\) or \(-5i\), with no real component involved. These are part of complex numbers, which are foundational in understanding advanced numbers, extending our number system beyond the reals.In our given quadratic equation, the concept of pure imaginary roots emerges due to the negative nature of \(-\frac{c}{a}\), leading us to the solutions:\[ x = \pm \sqrt{-\frac{c}{a}} \]This equation involves calculating the square root of a negative number, yielding imaginary numbers. Since there is no real part, these roots are called pure imaginary.

Sign of Coefficients

The sign of the coefficients in a quadratic equation plays a critical role in determining the nature of its roots. Here, we consider the signs of \(a\) and \(c\). When these coefficients are positive or negative, and they share the same sign, their effects compound in interesting ways.For example:

• If both \(a\) and \(c\) are positive, their quotient \(\frac{c}{a}\) is positive, making \(-\frac{c}{a}\) negative. This results in imaginary roots.
• If both \(a\) and \(c\) are negative, the same happens because their negatives cancel each other. This still yields a positive \(\frac{c}{a}\), making \(-\frac{c}{a}\) negative.

Thus, when \(a\) and \(c\) possess the same sign, \(-\frac{c}{a}\) is inherently negative, driving the quadratic equation towards pure imaginary roots. This is a fascinating outcome, as it veers the solutions from the real number spectrum into the conceptual realm of imaginary numbers.