Problem 54 - Algebra 2

Question

Which quadratic equation has roots \(\frac{1}{2}\) and \(\frac{1}{3} ?\) A. \(5 x^{2}-5 x-2=0\) B. \(5 x^{2}-5 x+1=0\) C. \(6 x^{2}+5 x-1=0\) D. \(6 x^{2}-5 x+1=0\)

Step-by-Step Solution

Answer

The correct quadratic equation is option D: \(6x^2 - 5x + 1 = 0\).

Step 1: Use Vieta’s Formulas for Sum of Roots

By Vieta's formulas, the sum of the roots of a quadratic equation \( ax^2 + bx + c = 0 \) is given by \(-b/a\). For the roots \(\frac{1}{2}\) and \(\frac{1}{3}\), the sum is \(\frac{1}{2} + \frac{1}{3} = \frac{3}{6} + \frac{2}{6} = \frac{5}{6}\). This means: \(-b/a = \frac{5}{6}\).

Step 2: Use Vieta’s Formula for Product of Roots

The product of the roots of a quadratic equation is given by \(c/a\). With roots \(\frac{1}{2}\) and \(\frac{1}{3}\), the product is \(\frac{1}{2} \times \frac{1}{3} = \frac{1}{6}\). This implies:\(c/a = \frac{1}{6}\).

Step 3: Identify the Correct Quadratic Equation

Given options are:- A: \(a=5, b=-5, c=-2\)- B: \(a=5, b=-5, c=1\)- C: \(a=6, b=5, c=-1\)- D: \(a=6, b=-5, c=1\)Use the values of \(-b/a\) and \(c/a\) calculated previously:- Option D: \(-b/a = -\frac{-5}{6} = \frac{5}{6}\) and \(c/a = \frac{1}{6}\).Option D meets both criteria from Vieta's formulas.

Key Concepts

Vieta's Formulas

Vieta's formulas are an essential tool for solving quadratic equations and understanding their properties. Named after the French mathematician François Viète, these formulas connect the coefficients of a polynomial equation with sums and products of its roots. In the context of a quadratic equation, which is an equation of the form \( ax^2 + bx + c = 0 \), Vieta's formulas provide a powerful shortcut for finding relationships without directly solving the equation.

• For the sum of the roots of a quadratic equation, Vieta's formula states that the sum is equal to \(-b/a\).
• The product of the roots, according to Vieta's formula, is \(c/a\).

By knowing these relationships, we can easily derive same-information needed in our equation to connect the formulation to its roots quickly. These formulas not only aid in finding missing coefficients but also ensure that if you know the roots directly, you can directly write down the equation.

Sum of Roots

The sum of the roots of a quadratic equation can be easily determined using Vieta's formula. For a standard quadratic equation \( ax^2 + bx + c = 0 \), the sum of roots \((r_1 + r_2)\) is given by \(-\frac{b}{a}\).
In practical terms, this means for any two roots \(r_1\) and \(r_2\), the sum \( r_1 + r_2 \) is simply the opposite of the ratio of the linear coefficient \(b\) to the leading coefficient \(a\). In the original exercise, with roots \(\frac{1}{2}\) and \(\frac{1}{3}\), the sum is \(\frac{5}{6}\), calculated as \(\frac{1}{2} + \frac{1}{3} = \frac{5}{6}\).
Understanding this concept helps in quickly matching equations with the given roots based on the sum alone.

Product of Roots

Another critical aspect of quadratic equations is the product of their roots, which can also be found using Vieta's formula. For a quadratic equation of the form \( ax^2 + bx + c = 0 \), the product of the roots \((r_1 \times r_2)\) is expressed as \(\frac{c}{a}\).
This formula directly links the product of the roots to the ratio of the constant term \(c\) and the leading coefficient \(a\). In the provided exercise, using the roots \(\frac{1}{2}\) and \(\frac{1}{3}\), the product is \(\frac{1}{6}\), as given by \(\frac{1}{2} \times \frac{1}{3} = \frac{1}{6}\).
By applying this formula, you can determine which quadratic equation matches the given product condition, offering an effective method to identify the correct polynomial even without direct factorization or graphing.