Problem 18 - College Algebra with Corequisite Support

Question

For the following exercises, solve the quadratic equation by factoring. $$ \frac{x}{3}-\frac{9}{x}=2 $$

Step-by-Step Solution

Answer

The solutions are \(x = 9\) and \(x = -3\).

Step 1: Eliminate the Fractions

First, we want to eliminate the fractions in the equation by multiplying both sides by the least common multiple of the denominators. The denominators are 3 and \(x\), so the LCM is \(3x\). Multiply each term by \(3x\): \[ \frac{x}{3} \cdot 3x - \frac{9}{x} \cdot 3x = 2 \cdot 3x \]This simplifies to:\[ x^2 - 27 = 6x \]

Step 2: Rearrange to Standard Form

Next, rearrange the equation into the standard form of a quadratic equation \(ax^2 + bx + c = 0\): \[ x^2 - 6x - 27 = 0 \]

Step 3: Factor the Quadratic Equation

Now, factor the quadratic equation. We need two numbers that multiply to -27 and add to -6. These numbers are -9 and 3. So, factor as follows:\[ (x - 9)(x + 3) = 0 \]

Step 4: Solve for x Using the Zero Product Property

Using the zero product property, set each factor equal to zero and solve for \(x\):\[\begin{align*} x - 9 &= 0 \ x &= 9 \ \ x + 3 &= 0 \ x &= -3\end{align*}\]

Step 5: Verify the Solutions

Always verify the solutions in the original equation. Plug \(x = 9\) and \(x = -3\) back into the original equation:For \(x = 9\):\[ \frac{9}{3} - \frac{9}{9} = 2 \rightarrow 3 - 1 = 2 \] (True)For \(x = -3\):\[ \frac{-3}{3} - \frac{9}{-3} = 2 \rightarrow -1 + 3 = 2 \] (True)Both solutions satisfy the original equation.

Key Concepts

Zero Product Property

The Zero Product Property is a fundamental concept in algebra. It states that if the product of two numbers is zero, then at least one of the numbers must be zero. This principle is particularly useful when solving quadratic equations that can be factored.

• This property tells us that for any two numbers, if \( a \cdot b = 0 \), then either \( a = 0 \), \( b = 0 \), or both.
• It allows us to break down factored quadratic equations into simpler linear equations that are easier to solve.

In the example \((x - 9)(x + 3) = 0\), applying the Zero Product Property means setting each factor, \(x - 9\) and \(x + 3\), equal to zero.
Solving these simple equations (\(x - 9 = 0\) and \(x + 3 = 0\)) gives us the solutions \(x = 9\) and \(x = -3\).

Standard Form of a Quadratic Equation

The standard form of a quadratic equation is a specific way of writing the equation so it’s easier to solve. It is written as \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants and \(a eq 0\).

• Getting an equation into this form helps in applying various solution methods such as factoring, completing the square, or using the quadratic formula.

Rearranging our example equation \(x^2 - 27 = 6x\) into standard form results in \(x^2 - 6x - 27 = 0\).
This rearrangement is a crucial step before solving by factoring, as it organizes the equation in a way that exposes patterns useful for finding its roots.

Least Common Multiple

The Least Common Multiple (LCM) is the smallest multiple that two or more numbers share. It's especially useful for eliminating fractions from equations.

• In equations, the LCM helps clear denominators by making it possible to multiply through by a single number.

For instance, in the original equation \(\frac{x}{3} - \frac{9}{x} = 2\), the denominators are the numbers and variable 3 and \(x\). The LCM here is \(3x\).
Multiplying each term by \(3x\) eliminates the fractions, transforming the equation into \(x^2 - 27 = 6x\). This process simplifies the equation, paving the way for easier manipulation and solution.

Quadratic Equation

Quadratic equations are polynomial equations of the second degree, typically represented as \(ax^2 + bx + c = 0\). They have various properties:

• They can have two, one, or no real solutions.
• The solutions can be found using methods such as factoring, graphing, completing the square, or the quadratic formula.

Solving a quadratic by factoring involves rewriting it as a product of binomials. In our case, the equation \(x^2 - 6x - 27 = 0\) is factored into \((x - 9)(x + 3) = 0\).
Such factoring simplifies solving the equation by using the Zero Product Property. Factoring is often the simplest method if the quadratic is easily factorable.

Verify Solutions in Original Equation

Verifying solutions is an important step that ensures the accuracy of your results. After solving the equation, plug the solutions back into the original equation to see if they satisfy it.

• This step prevents any errors that might have crept in during the manipulation or simplification of the equation.
• It assures that the solutions obtained are true for the equation given initially, not just the modified form.

For our solutions, \(x = 9\) and \(x = -3\), substituting them back into the original equation \(\frac{x}{3} - \frac{9}{x} = 2\) showed both to be correct.
Upon substitution, both lead to a true statement: for \(x = 9\), it returns \(2\), and for \(x = -3\), it also returns \(2\). This verification confirms the solutions are indeed correct and satisfy the original equation.