Problem 116 - Elementary Algebra

Question

Solve by using the Quadratic Formula. \(\frac{1}{3} n^{2}+n=-\frac{1}{2}\)

Step-by-Step Solution

Answer

The solutions are \[ n = -\frac{3}{2} + \frac{\sqrt{3}}{2} \] and \[ n = -\frac{3}{2} - \frac{\sqrt{3}}{2} \].

Step 1: Standard Form of Quadratic Equation

The quadratic formula can only be applied if the quadratic equation is in the standard form: \[ ax^2 + bx + c = 0 \] First, bring the given equation into that form by moving all terms to one side: \[ \frac{1}{3} n^2 + n + \frac{1}{2} = 0 \]

Step 2: Identify Coefficients

Compare the equation \[ \frac{1}{3} n^2 + n + \frac{1}{2} = 0 \] with the standard form \[ ax^2 + bx + c = 0 \] to identify the coefficients. Here: \[ a = \frac{1}{3}, \quad b = 1, \quad c = \frac{1}{2} \]

Step 3: Apply the Quadratic Formula

The quadratic formula is given by: \[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Substitute the identified coefficients \(a, b, c\) into the formula.

Step 4: Calculate the Discriminant

Calculate the discriminant \( \Delta \), which is \( b^2 - 4ac \). Substitute the values: \[ \Delta = 1^2 - 4 \left( \frac{1}{3} \right) \left( \frac{1}{2} \right) \] \[ \Delta = 1 - \frac{2}{3} = \frac{3}{3} - \frac{2}{3} = \frac{1}{3} \]

Step 5: Solve for n

Use the quadratic formula to find the roots: \[ n = \frac{-1 \pm \sqrt{\frac{1}{3}}}{2 \cdot \frac{1}{3}} \] First, simplify the expression under the square root: \[ \sqrt{\frac{1}{3}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3} \] Then, simplify the denominator: \[ 2 \cdot \frac{1}{3} = \frac{2}{3} \] So, the expression becomes: \[ n = \frac{-1 \pm \frac{\sqrt{3}}{3}}{\frac{2}{3}} \]

Step 6: Simplify the Roots

Divide by \( \frac{2}{3} \): \[ n = \left( -1 \pm \frac{\sqrt{3}}{3} \right) \div \frac{2}{3} = \left( -1 \pm \frac{\sqrt{3}}{3} \right) \times \frac{3}{2} \] This results in: \[ n = -1 \times \frac{3}{2} \pm \frac{\sqrt{3}}{3} \times \frac{3}{2} \] \[ n = -\frac{3}{2} \pm \frac{\sqrt{3}}{2} \]

Key Concepts

Standard Form of Quadratic Equation

Before using the quadratic formula, it's important to make sure the equation is in standard form. The standard form of a quadratic equation is represented as: \[ ax^2 + bx + c = 0 \] In this form, \(a \), \( b \), and \( c \) are known as coefficients. If the original equation is not in this form, you need to rearrange it by moving all terms to one side of the equation. For example, in the given exercise, the original equation is: \[ \frac{1}{3} n^2 + n = -\frac{1}{2} \] To reformat this into the standard form, you should add \( \frac{1}{2} \) to both sides: \[ \frac{1}{3} n^2 + n + \frac{1}{2} = 0 \] Now, the equation is in standard form and ready for the application of the quadratic formula.

Quadratic Equation Coefficients

Coefficients are the numerical constants that multiply the variables in a polynomial. For the equation in standard form \[ ax^2 + bx + c = 0 \], the terms \(a\), \(b\), and \(c\) are the coefficients. In our example: \[ \frac{1}{3} n^2 + n + \frac{1}{2} = 0 \] By comparing this with \[ ax^2 + bx + c = 0 \], we can identify the coefficients as follows: - \(a\) is the coefficient of \(n^2\) and here \(a = \frac{1}{3}\) - \(b\) is the coefficient of \(n\) and here \(b = 1\) - \(c\) is the constant term and here \(c = \frac{1}{2}\) Understanding these coefficients is crucial, as they are used in the quadratic formula to find the roots of the equation.

Discriminant Calculation

The discriminant of a quadratic equation is a part of the quadratic formula that gives important information about the nature of the roots. The discriminant is represented by the symbol \( \Delta \), and it is calculated using the following formula: \[ \Delta = b^2 - 4ac \] For our given equation, the coefficients are: \(a = \frac{1}{3}\), \(b = 1\), and \(c = \frac{1}{2}\). Substituting these into the discriminant formula: \[ \Delta = 1^2 - 4 \left( \frac{1}{3} \right) \left( \frac{1}{2} \right) \] Calculate each term individually: \[ \Delta = 1 - \frac{2}{3} \] Simplify the result: \[ \Delta = \frac{3}{3} - \frac{2}{3} = \frac{1}{3} \] A positive discriminant here indicates that the quadratic equation has 2 real and distinct roots.

Roots of Quadratic Equation

The roots of a quadratic equation are the solutions to the equation \[ ax^2 + bx + c = 0 \]. These roots can be found using the quadratic formula: \[ n = \frac{-b \pm \sqrt{\Delta}}{2a} \] Substituting the values of \(a, b,\) and \( \Delta \) into the formula, we get: \[ n = \frac{-1 \pm \sqrt{\frac{1}{3}}}{2 \cdot \frac{1}{3}} \] Simplify the square root and the fraction under the root: \[ \sqrt{\frac{1}{3}} = \frac{1}{\sqrt{3}} = \frac{ \sqrt{3}}{3} \] Simplify the denominator: \[ 2 \cdot \frac{1}{3} = \frac{2}{3} \] Now the equation becomes: \[ n = \frac{-1 \pm \frac{ \sqrt{3}}{3}}{ \frac{2}{3}} \] To divide by \( \frac{2}{3} \), multiply by \( \frac{3}{2} \): \[ n = \left( -1 \pm \frac{ \sqrt{3}}{3} \right) \div \frac{2}{3} = \left( -1 \pm \frac{ \sqrt{3}}{3} \right) \times \frac{3}{2} \] Simplify further: \[ n = -\frac{3}{2} \pm \frac{ \sqrt{3}}{2} \] These are the roots: \[ n = -\frac{3}{2} + \frac{ \sqrt{3}}{2} \] and \[ n = -\frac{3}{2} - \frac{ \sqrt{3}}{2} \]