Problem 40 - Algebra 2 and Trigonometry

Question

Write a quadratic equation with integer coefficients for each pair of roots. \(\frac{-1+\sqrt{3}}{3}, \frac{-1-\sqrt{3}}{3}\)

Step-by-Step Solution

Answer

The quadratic equation is \(3x^2 + 2x - 2 = 0\).

Step 1: Understand the Roots

The given roots are \( r_1 = \frac{-1 + \sqrt{3}}{3} \) and \( r_2 = \frac{-1 - \sqrt{3}}{3} \). These roots are derived from factoring a quadratic equation of the form \( ax^2 + bx + c = 0 \).

Step 2: Form the Quadratic Equation

A quadratic equation with roots \( r_1 \) and \( r_2 \) can be expressed using the formula \( (x - r_1)(x - r_2) = 0 \). This expands to \( x^2 - (r_1 + r_2)x + r_1r_2 = 0 \).

Step 3: Calculate the Sum and Product of Roots

First, find the sum of the roots:\[r_1 + r_2 = \frac{-1 + \sqrt{3}}{3} + \frac{-1 - \sqrt{3}}{3} = \frac{-1 + \sqrt{3} - 1 - \sqrt{3}}{3} = \frac{-2}{3}\]Next, find the product of the roots:\[r_1r_2 = \left(\frac{-1 + \sqrt{3}}{3}\right) \left(\frac{-1 - \sqrt{3}}{3}\right) = \frac{(-1 + \sqrt{3})(-1 - \sqrt{3})}{9} = \frac{1 - 3}{9} = \frac{-2}{9}\]

Step 4: Substitute Into the Quadratic Formula

Substitute the sum and product of the roots into the standard form of a quadratic equation: \[x^2 - \left(\frac{-2}{3}\right)x + \frac{-2}{9} = 0\] This simplifies to:\[3x^2 + 2x - 2 = 0\] by multiplying through by 9 to clear the fractions.

Key Concepts

Roots of Equations

The roots of an equation are the values of the variable that satisfy the equation, essentially making the equation equal to zero. In the case of quadratic equations, these roots are what make the expression \( ax^2 + bx + c = 0 \) true.
Quadratic equations have a special property: they always have at most two roots. These can be real or complex numbers.

• Real roots occur when the graph of the quadratic function touches or crosses the x-axis.
• Complex roots, which occur in conjugate pairs, do not intersect the x-axis and are often expressed with square roots of negative numbers.

For our given roots, \( r_1 = \frac{-1 + \sqrt{3}}{3} \) and \( r_2 = \frac{-1 - \sqrt{3}}{3} \), both are real and derived from using the quadratic formula.

Integer Coefficients

When crafting a quadratic equation from given roots, especially fractions or irrational numbers, it's crucial to convert it into a format with integer coefficients. This is important in mathematics since it yields simpler equations that are easier to work with and understand.
The transformation involves clearing any fractions by multiplying all terms by the least common denominator (LCD).
In our example, we started with roots that implied fractional coefficients in the basic quadratic expression. By multiplying through by 9, the LCD here, we cleared the fractions, resulting in the following equation with integer coefficients: \[3x^2 + 2x - 2 = 0\].
Ensuring integer coefficients in any polynomial equation not only simplifies the calculation process but also adheres to standard forms used in higher-level mathematics.

Sum and Product of Roots

The sum and product of the roots of a quadratic equation provide insightful relationships applied often in algebra without solving the full equation.
For a quadratic equation in the form \(ax^2 + bx + c = 0\), the relationships are:

• Sum of roots, \( r_1 + r_2 = -\frac{b}{a} \)
• Product of roots, \( r_1r_2 = \frac{c}{a} \)

These relationships derive directly from expanding the product \((x - r_1)(x - r_2)\) and observing the coefficients that emerge.
Using the provided roots, \( r_1 = \frac{-1 + \sqrt{3}}{3} \) and \( r_2 = \frac{-1 - \sqrt{3}}{3} \):
- The sum is \(\frac{-2}{3}\), giving coefficient \(b\) a value such that \(-\frac{b}{3} = \frac{-2}{3} \), leading to \( b = 2 \).- The product is \(\frac{-2}{9}\), thus \( \frac{c}{3} = \frac{-2}{9} \), leading to \( c = -2 \).This information helps rebuild the quadratic equation efficiently from the roots without solving the quadratic formula forward again.