Chapter 11: Problem 14
Use the quadratic formula to solve each equation. These equations have real number solutions only. See Examples I through 3. $$ \frac{1}{6} x^{2}+x+\frac{1}{3}=0 $$
Short Answer
Expert verified
The solutions are \( x = -3 + \sqrt{7} \) and \( x = -3 - \sqrt{7} \).
Step by step solution
01
Identify coefficients
The given equation is \( \frac{1}{6}x^2 + x + \frac{1}{3} = 0 \). First, we need to identify the coefficients \( a \), \( b \), and \( c \) where the general form of a quadratic equation is \( ax^2 + bx + c = 0 \). Here, \( a = \frac{1}{6} \), \( b = 1 \), and \( c = \frac{1}{3} \).
02
Write down the quadratic formula
The quadratic formula is given by \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). This formula will be used to find the roots of the equation.
03
Calculate the discriminant
The discriminant \( D \) is \( b^2 - 4ac \). Plug in the coefficients: \( b = 1 \), \( a = \frac{1}{6} \), \( c = \frac{1}{3} \). So, we compute: \( D = 1^2 - 4 \times \frac{1}{6} \times \frac{1}{3} = 1 - \frac{4}{18} = 1 - \frac{2}{9} = \frac{9}{9} - \frac{2}{9} = \frac{7}{9} \). The discriminant is \( \frac{7}{9} \), which is positive, confirming real solutions.
04
Substitute into the quadratic formula
Substitute the values into the quadratic formula: \( x = \frac{-1 \pm \sqrt{\frac{7}{9}}}{\frac{2}{6}} \). Simplify the terms: \( x = \frac{-1 \pm \frac{\sqrt{7}}{3}}{\frac{1}{3}} \).
05
Simplify the solution
Multiply numerator and denominator by 3 to simplify: \( x = \frac{-1 \pm \frac{\sqrt{7}}{3}}{\frac{1}{3}} = \frac{-3 \pm \sqrt{7}}{1} = -3 \pm \sqrt{7} \). Therefore, the solutions are \( x_1 = -3 + \sqrt{7} \) and \( x_2 = -3 - \sqrt{7} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quadratic Equations
Quadratic equations are mathematical expressions of the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are coefficients, and \( a eq 0 \). These equations are used to model various real-life scenarios, ranging from calculating areas to predicting outcomes in physics and economics. Quadratics are characterized by the highest exponent of the variable \( x \) being 2, and this distinct form results in a parabolic graph when plotted on a coordinate plane.
Understanding the components:
Understanding the components:
- \( a \) - the coefficient of \( x^2 \), which affects the "opening" direction and width of the parabola.
- \( b \) - the coefficient of \( x \), which influences the parabola's position along the x-axis.
- \( c \) - the constant term, which dictates the point where the parabola intersects the y-axis.
Discriminant
The discriminant is a vital part of solving quadratic equations as it helps us understand the nature of the solutions. For any quadratic equation \( ax^2 + bx + c = 0 \), the discriminant \( D \) is given by the expression \( b^2 - 4ac \). This value determines how many and what type of solutions the quadratic equation has.
Let's break down its significance:
Let's break down its significance:
- A positive discriminant (\( D > 0 \)) indicates two distinct real number solutions.
- A zero discriminant (\( D = 0 \)) reveals there is exactly one real solution, also called a repeated or double root.
- A negative discriminant (\( D < 0 \)) shows there are no real solutions, but instead two complex solutions.
Real Number Solutions
Real number solutions for quadratic equations are the roots where the graph of a quadratic equation intersects the x-axis. These solutions can be found using the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). This formula provides the tool for calculating these roots when the quadratic equation resists simpler solving methods like factoring.
When using the quadratic formula:
When using the quadratic formula:
- Insert the coefficients \( a \), \( b \), and \( c \) into the formula.
- Compute the discriminant \( b^2 - 4ac \) to determine the type of roots.
- Simplify under the square root to find the exact values of the solutions.