Problem 2

Question

Write a quadratic equation with the given root(s). Write the equation in standard form. \(\frac{1}{2}, \frac{4}{3}\)

Step-by-Step Solution

Verified
Answer
The quadratic equation is \( 6x^2 - 11x + 2 = 0 \).
1Step 1: Understand the Relationship of Roots to Factors
The roots of a quadratic equation are related to its factors. If the roots are \( r_1 \) and \( r_2 \), then the quadratic equation can be expressed as \( (x - r_1)(x - r_2) = 0 \). In this case, our roots are \( \frac{1}{2} \) and \( \frac{4}{3} \).
2Step 2: Write the Factors using Given Roots
Using the roots \( \frac{1}{2} \) and \( \frac{4}{3} \), the factors of the equation are \( (x - \frac{1}{2}) \) and \( (x - \frac{4}{3}) \).
3Step 3: Form the Quadratic Equation from the Factors
Multiply the factors to form the quadratic equation: \[ (x - \frac{1}{2})(x - \frac{4}{3}) = 0 \].
4Step 4: Expand the Expression
Use the distributive property (FOIL method) to expand: \[ (x - \frac{1}{2})(x - \frac{4}{3}) = x^2 - \frac{4}{3}x - \frac{1}{2}x + \frac{2}{6} \].
5Step 5: Simplify the Equation
Combine like terms: \[ x^2 - (\frac{4}{3}x + \frac{1}{2}x) + \frac{1}{3} \].Convert \( \frac{4}{3}x \) to have the same denominator as \( \frac{1}{2}x \).\[ \frac{4}{3}x = \frac{8}{6}x, \quad \frac{1}{2}x = \frac{3}{6}x \], the expression becomes \[ x^2 - \frac{11}{6}x + \frac{1}{3} \].
6Step 6: Convert to Standard Form
Multiply the whole equation by 6 to eliminate fractions: \[ 6(x^2 - \frac{11}{6}x + \frac{1}{3}) = 0 \]Simplifying gives: \[ 6x^2 - 11x + 2 = 0 \].This is the quadratic equation in standard form \( ax^2 + bx + c = 0 \).

Key Concepts

Roots of Quadratic EquationsStandard Form of Quadratic EquationsFOIL MethodDistributive Property in Algebra
Roots of Quadratic Equations
Quadratic equations often have two solutions or roots. These roots can be expressed using the formula associated with the quadratic equation. If you're given specific roots, such as \(\frac{1}{2}\) and \(\frac{4}{3}\), you can write the equation as:
  • \((x - \text{root}_1)(x - \text{root}_2) = 0\)
In our example:
  • \((x - \frac{1}{2})(x - \frac{4}{3}) = 0\)
From this expression, you can find the original quadratic equation by reversing the process used to solve it.
Standard Form of Quadratic Equations
A quadratic equation in its simplest form is written as \(ax^2 + bx + c = 0\). This structure is called the standard form.
To convert an equation into this form, ensure:
  • All terms are collected on one side of the equation.
  • Coefficients \(a\), \(b\), and \(c\) are integers, if possible.
For instance, after expanding and simplifying the equation from roots \(\frac{1}{2}\) and \(\frac{4}{3}\), the final standard form is \(6x^2 - 11x + 2 = 0\). This ensures the equation is clear and ready for analysis.
FOIL Method
The FOIL method is a technique used to expand two binomials. FOIL stands for First, Outer, Inner, Last, representing:
  • First: Multiply the first terms of each binomial.
  • Outer: Multiply the outer terms.
  • Inner: Multiply the inner terms.
  • Last: Multiply the last terms of each binomial.
In the example of \((x - \frac{1}{2})(x - \frac{4}{3})\):
  • First: \(x \cdot x = x^2\)
  • Outer: \(x \cdot (-\frac{4}{3}) = -\frac{4}{3}x\)
  • Inner: \((-\frac{1}{2}) \cdot x = -\frac{1}{2}x\)
  • Last: \((-\frac{1}{2}) \cdot (-\frac{4}{3}) = \frac{2}{6}\)
Combine these results for the expanded expression, which helps in equation formation and simplification.
Distributive Property in Algebra
The distributive property helps in multiplying a single term by two or more terms inside a bracket: \(a(b + c) = ab + ac\). This property is extended to distributing across subtraction and division as well.
In dealing with quadratic equations, the distributive property becomes essential when we expand expressions. When applied to our example \(\left(x - \frac{1}{2}\right)\left(x - \frac{4}{3}\right)\), each term inside the first parenthesis is multiplied with each term inside the second parenthesis, as shown through FOIL.
Applying this property ensures that we don't miss any terms and aids in finding the correct expression during multiplication, making it vital for forming accurate quadratic equations.