Problem 128 - Elementary Algebra

Question

Solve by using the Quadratic Formula. \(\frac{1}{9} c^{2}+\frac{2}{3} c=3\)

Step-by-Step Solution

Answer

c = 3 or c = -9

Step 1 - Convert the Equation to Standard Form

First, convert the given equation \(\frac{1}{9} c^{2}+\frac{2}{3} c=3\) into the standard quadratic form \(ax^2 + bx + c = 0\). To do this, subtract 3 from both sides.\(\frac{1}{9} c^{2}+\frac{2}{3} c - 3 = 0\)

Step 2 - Clear the Fraction Coefficients

Next, clear the fractions by multiplying every term by 9 (the least common multiple of 9) to obtain integer coefficients:\(1c^2 + 6c - 27 = 0\) or simply \c^2 + 6c - 27 = 0\.

Step 3 - Identify the Coefficients a, b, and c

The quadratic formula is \(c = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). In this equation, identify \(a = 1\), \(b = 6\), and \(c = -27\).

Step 4 - Plug Coefficients into the Quadratic Formula

Substitute \(a, b,\) and \(c\) into the quadratic formula:\(c = \frac{-6 \pm \sqrt{6^2 - 4 \cdot 1 \cdot (-27)}}{2 \cdot 1}\)

Step 5 - Simplify Under the Square Root

Calculate the value inside the square root:\(c = \frac{-6 \pm \sqrt{36 + 108}}{2}\) which simplifies to \(c = \frac{-6 \pm \sqrt{144}}{2}\).

Step 6 - Solve for the Roots

Simplify further to find the roots:\(c = \frac{-6 \pm 12}{2}\). This gives two solutions: \(c = \frac{-6 + 12}{2} = 3\) and \(c = \frac{-6 - 12}{2} = -9\).

Key Concepts

Convert to Standard Form

To solve a quadratic equation using the Quadratic Formula, we first need to convert it into the standard form \(ax^2 + bx + c = 0\). The given equation is \(\frac{1}{9} c^{2} + \frac{2}{3} c = 3\). To convert this to standard form, we need to move all terms to one side of the equation. Subtract 3 from both sides of the equation: \(\frac{1}{9} c^{2} + \frac{2}{3} c - 3 = 0\).
This step transforms the equation into a format that allows for further simplification and sets us up for using the Quadratic Formula effectively.

Clear Fraction Coefficients

Having an equation with fractions makes calculations cumbersome. To clear the fraction coefficients, we need to multiply every term by the least common multiple (LCM) of the denominators. Here, the denominators are 9 and 3, and their LCM is 9.
So we multiply each term in the equation \(\frac{1}{9} c^{2} + \frac{2}{3} c - 3 = 0\) by 9:
\9 \times \frac{1}{9} c^2 + 9 \times \frac{2}{3} c - 9 \times 3 = 0\ which simplifies to \c^2 + 6c - 27 = 0\.
By clearing the fractions, we obtain integer coefficients, making the equation cleaner and easier to work with using the Quadratic Formula.

Identify Coefficients

In the quadratic equation we derived \(c^2 + 6c - 27 = 0\), it is essential to identify the coefficients a, b, and c correctly. The standard form \(ax^2 + bx + c = 0\) allows for easy identification of these coefficients:

• The coefficient \(a\) is the term multiplying the squared variable, so here \(a = 1\).
• The coefficient \(b\) is the term multiplying the variable, so here \(b = 6\).
• The constant term \(c\) is the standalone number, so here \(c = -27\).

With these coefficients identified, we can now substitute them into the Quadratic Formula: \x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\. Identifying coefficients correctly is crucial for solving the equation accurately.