Problem 19 - Brief Applied Calculus

Question

Solve each equation by factoring. [Hint for Exer cises 19-22: First factor out a fractional power.] $$ 3 x^{5 / 2}-6 x^{3 / 2}=9 x^{1 / 2} $$

Step-by-Step Solution

Answer

The solutions are \(x = 0\), \(x = 3\), and \(x = -1\).

Step 1: Identify a Common Factor

Examine the equation and identify the common factor in all the terms. Here, all terms contain a factor of \(x^{1/2}\). Factoring this out, you get: \[ x^{1/2}(3x^2 - 6x - 9) = 0 \] This simplifies the equation and allows us to solve it more easily.

Step 2: Solve for Zero Product Property

According to the Zero Product Property, if the product of two factors is zero, then at least one of the factors must be zero. The expression inside the parenthesis can be set to zero: \[ 3x^2 - 6x - 9 = 0 \] and also the common factor: \[ x^{1/2} = 0 \] Now, solve these two equations separately.

Step 3: Solve for \(x\) from \(x^{1/2}=0\)

For \(x^{1/2} = 0\), square both sides to eliminate the fractional exponent: \[ x = 0 \]This gives one solution for the equation.

Step 4: Factor the Quadratic Equation

For the equation \(3x^2 - 6x - 9 = 0\), first factor out the greatest common factor, which is 3: \[ 3(x^2 - 2x - 3) = 0 \] Simplifying gives: \[ x^2 - 2x - 3 = 0 \] We then factor the quadratic expression.

Step 5: Factor the Quadratic Expression

Look for two numbers that multiply to \(-3\) and add to \(-2\). The numbers \(-3\) and \(1\) work: \[ (x - 3)(x + 1) = 0 \] This gives the two potential solutions for \(x\): \(x-3=0\) and \(x+1=0\).

Step 6: Solve for \(x\) from Factored Expression

From the factors \(x-3=0\) and \(x+1=0\), solve each for \(x\):- For \(x - 3 = 0\), add 3 to both sides: \[ x = 3 \]- For \(x + 1 = 0\), subtract 1 from both sides:\[ x = -1 \]Therefore, the solutions are \(x = 3\) and \(x = -1\).

Key Concepts

Fractional Exponents

Fractional exponents are a way to express roots using exponent notation. They are particularly useful for simplifying and solving equations. The exponent of 1/2, for example, corresponds to taking the square root of a number.

• The expression \( x^{1/2} \) means the square root of \( x \).
• The expression \( x^{m/n} \) represents the n-th root of \( x \) raised to the power of \( m \).

Fractional exponents turn a root operation into something we can handle with our normal rules for exponents, making them easier to manage in algebraic expressions. Let's take an example from our exercise:
In the equation \( 3x^{5/2} - 6x^{3/2} = 9x^{1/2} \), each term contains a fractional exponent. To solve the equation, we factor out the common term \( x^{1/2} \). This is similar to finding a common factor in integer exponents, which simplifies the equation and makes it easier to solve.

Zero Product Property

The Zero Product Property is a fundamental principle in algebra. It states that if the product of two or more factors is zero, then at least one of the factors must be zero. This property is incredibly helpful when solving polynomial equations.

• If \( ab = 0 \), then either \( a = 0 \) or \( b = 0 \) (or both).

In the exercise, after factoring out \( x^{1/2} \), the equation was reduced to \( x^{1/2}(3x^2 - 6x - 9) = 0 \). According to the Zero Product Property, we then separate and solve each part:

• For \( x^{1/2} = 0 \), squaring both sides gives \( x = 0 \).
• For \( 3x^2 - 6x - 9 = 0 \), further simplification and factoring are needed.

This property simplifies the process of finding solutions, by breaking down complex equations into smaller, simpler pieces.

Quadratic Formula

The Quadratic Formula is a powerful tool for solving any quadratic equation of the form \( ax^2 + bx + c = 0 \). Even when factoring isn't feasible, the formula provides the solutions directly:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
This equation gives the values of \( x \) that satisfy the quadratic equation. The term under the square root, called the discriminant (\( b^2 - 4ac \)), determines the nature of the roots:

• If the discriminant is positive, there are two distinct real roots.
• If it is zero, there is one real root (also called a double root).
• If it is negative, the roots are complex or imaginary.

In the exercise solution, once we reached the simpler quadratic equation \( x^2 - 2x - 3 = 0 \), it was able to be factored easily. However, if it hadn't been factorable directly, the Quadratic Formula would be our go-to method for an accurate solution.