Problem 20 - Applied Calculus

Question

Solve each equation by factoring. [Hint for: First factor out a fractional power.] $$ 2 x^{7 / 2}+8 x^{5 / 2}=24 x^{3 / 2} $$

Step-by-Step Solution

Answer

The solutions are \(x = 0, x = -6,\) and \(x = 2\).

Step 1: Identify the common factor

Examine the given equation \(2x^{7/2} + 8x^{5/2} = 24x^{3/2}\). Notice that every term has a common factor of \(x^{3/2}\). We will factor this common term out from each part of the equation.

Step 2: Factor out the common power

Take \(x^{3/2}\) as a common factor from each term:\[2x^{7/2} + 8x^{5/2} - 24x^{3/2} = 0\] This becomes:\[x^{3/2}(2x^2 + 8x - 24) = 0\]

Step 3: Solve for zero factor product

For a product to be zero, at least one of the factors must be zero. Therefore, solve each factor separately. Start with \(x^{3/2} = 0\), which gives \(x = 0\). Now, factor the quadratic term \(2x^2 + 8x - 24 = 0\).

Step 4: Factor the quadratic expression

To factor \(2x^2 + 8x - 24\), first observe if any simplification is possible. We can divide the entire expression by 2:\[x^2 + 4x - 12 = 0\]Find two numbers that multiply to \(-12\) and add to \(4\). The numbers 6 and \(-2\) work, so use them to factor:\[(x + 6)(x - 2) = 0\].

Step 5: Solve for the roots of the quadratic equation

The factored quadratic expression \((x + 6)(x - 2) = 0\) provides solutions when each factor is set to zero:- \(x + 6 = 0\) implies \(x = -6\)- \(x - 2 = 0\) implies \(x = 2\).

Key Concepts

Factoring Polynomial Expressions

Factoring polynomial expressions refers to the process of expressing a polynomial as the product of its factors. Factors are simpler polynomials or numbers when multiply together give us the original polynomial.
To factor a polynomial, you typically need to identify its common factors or use specific factoring techniques such as grouping or using special formulas.

In the context of our exercise, we focused on identifying a common power in polynomial terms. We looked at an algebraic expression with fractional exponents and factored out the lowest common term. This stripped down the equation to a simpler form, making it easier to solve. Here's a simple way to approach factoring polynomial expressions:

• Identify the common factor in every term.
• Factor it out from each term to simplify the expression.
• Carefully factor any resulting quadratic or simpler expressions.

After factoring, each expression can be set to zero to find the roots of the polynomial. This is crucial for solving equations by finding the values that satisfy the equation set equal to zero.

Quadratic Equations

Quadratic equations are polynomials of degree two, meaning the highest exponent of the variable is two. They take the general form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants.
Solving quadratic equations usually involves factoring, using the quadratic formula, or completing the square.

In our exercise, after factoring out the fractional power, we ended up with a quadratic expression \(2x^2 + 8x - 24\). We simplified this by factoring, leading us to an expression \((x + 6)(x - 2) = 0\). Each factor gives solutions when it's set to zero.

Here are steps to keep in mind when solving quadratic equations:

• Ensure the equation is in standard quadratic form.
• Factor the quadratic if possible.
• Set each factor to zero and solve for the variable.

This method of solving makes equations manageable and easy to solve by breaking them into smaller, simpler problems.

Fractional Exponents

Fractional exponents might seem complex, but they offer a useful way to express roots.
For example, \(x^{1/2}\) is equivalent to the square root of \(x\). Thus, a fractional exponent represents the power and root of a number.

In the exercise, the equation \(2 x^{7 / 2}+8 x^{5 / 2}=24 x^{3 / 2}\) was given. Here, each term had a fractional exponent, which were rational numbers indicating both power and a root.
To solve the equation, it was crucial to factor out the lowest power, \(x^{3/2}\), common to all terms.

Understanding fractional exponents is beneficial, as it helps in:

• Simplifying algebraic expressions that involve roots.
• Applying laws of exponents to simplify complex calculations.
• Converting radical expressions into a more manageable form.

By mastering this concept, you gain flexibility in handling equations that mix powers and roots efficiently.