Problem 44 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. \(31-44\) : Solve by using the quadratic formula. $$ \frac{1}{7} x^{2}+1=\frac{4}{7} x $$

Step-by-Step Solution

Answer

The equation has no real solutions; it has complex solutions: \( x = 2 \pm i \frac{\sqrt{12}}{7} \).

Step 1: Write the Equation in Standard Form

The given quadratic equation is \( \frac{1}{7} x^{2}+1=\frac{4}{7} x \). Start by rearranging the terms to one side of the equation to obtain the standard form \( ax^2 + bx + c = 0 \). Subtract \( \frac{4}{7}x \) from both sides to get \( \frac{1}{7}x^2 - \frac{4}{7}x + 1 = 0 \).

Step 2: Identify Coefficients

Identify the coefficients \( a \), \( b \), and \( c \) from the standard form of the equation \( \frac{1}{7}x^2 - \frac{4}{7}x + 1 = 0 \). Here, \( a = \frac{1}{7} \), \( b = -\frac{4}{7} \), and \( c = 1 \).

Step 3: Write the Quadratic Formula

The quadratic formula is given by \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Use this formula to find the solutions for \( x \).

Step 4: Calculate the Discriminant

The discriminant, \( \Delta \), is calculated as \( b^2 - 4ac \). Substitute the identified values: \( (-\frac{4}{7})^2 - 4(\frac{1}{7})(1) = \frac{16}{49} - \frac{4}{7} \). Simplify by converting \( \frac{4}{7} \) to \( \frac{28}{49} \), so \( \Delta = \frac{16}{49} - \frac{28}{49} = -\frac{12}{49} \).

Step 5: Interpret the Discriminant Result

Since the discriminant \( \Delta = -\frac{12}{49} \) is negative, this indicates that there are no real solutions to the equation. The solutions will be complex conjugates.

Step 6: Compute Complex Solutions

Even though there are no real solutions, compute the complex solutions using the quadratic formula. Substitute \( a \), \( b \), \( c \), and \( \Delta \):\[ x = \frac{-(-\frac{4}{7}) \pm \sqrt{-\frac{12}{49}}}{2(\frac{1}{7})} \]. Further simplifying gives: \[ x = \frac{\frac{4}{7} \pm \sqrt{-\frac{12}{49}}}{\frac{2}{7}} \] or \[ x = 2 \pm i \frac{\sqrt{12}}{//7} \], representing the two complex conjugate solutions.

Key Concepts

Standard Form of Quadratic Equation

The standard form of a quadratic equation is crucial for utilizing the quadratic formula effectively. It is expressed as \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( a eq 0 \). This format organizes the equation properly, allowing you to correctly identify the coefficients necessary for solving it.

To convert the given equation \( \frac{1}{7} x^{2}+1=\frac{4}{7} x \) into its standard form, we rearrange all terms to one side so the equation becomes \( \frac{1}{7} x^2 - \frac{4}{7} x + 1 = 0 \).

• Step 1: Identify \( a \), \( b \), and \( c \). Here, \( a = \frac{1}{7} \), \( b = -\frac{4}{7} \), and \( c = 1 \).
• Step 2: Recognize the importance of this form: It sets the stage for applying methods like the quadratic formula or factoring methods.

Understanding the standard form helps break down complex equations into something manageable, paving the way for deeper analysis like solving for roots.

Discriminant

The discriminant is a powerful tool within the quadratic formula that helps us determine the nature of the roots of a quadratic equation without even solving it. It is calculated as \( \Delta = b^2 - 4ac \).

This simple calculation gives you immediate insight:

• If \( \Delta > 0 \), there are two distinct real solutions.
• If \( \Delta = 0 \), there is exactly one real solution (or a repeated root).
• If \( \Delta < 0 \), the solutions are complex.

In our equation, by substituting the coefficients into the formula, we find \( \Delta = (-\frac{4}{7})^2 - 4(\frac{1}{7})(1) = \frac{16}{49} - \frac{28}{49} = -\frac{12}{49} \).

Because the discriminant is negative, it indicates the absence of real solutions. Instead, the equation will have complex solutions, which are conjugates of each other, adding an interesting layer to the analysis.

Complex Solutions

When the discriminant of a quadratic equation is negative, as in the case of our equation with \( \Delta = -\frac{12}{49} \), the solutions are not real numbers but instead complex numbers. Complex solutions come in conjugate pairs and are expressed in the form \( a \pm bi \), where \( i \) is the imaginary unit defined by \( i^2 = -1 \).

Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) and computing it with our values, the solution can be expressed as:

• Find the real part: \( -b \) simplifies to \( \frac{4}{7} \).
• Find the imaginary part: \( \sqrt{-\frac{12}{49}} \) simplifies to \( i \frac{\sqrt{12}}{7} \).

This yields the complex solutions \( x = 2 \pm i \frac{\sqrt{12}}{7} \). These complex roots show that for the given equation, although real solutions are not present, complex solutions provide a nuanced understanding of the equation's behavior.