Problem 21 - Applied Calculus

Question

Solve each equation by factoring. [Hint for: First factor out a fractional power.] $$ 2 x^{5 / 2}+4 x^{3 / 2}=6 x^{1 / 2} $$

Step-by-Step Solution

Answer

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

Step 1: Identify Common Factor

Let's start by identifying the common factor in the equation. Notice that each term contains \(x^{1/2}\). Thus, \(x^{1/2}\) is a common factor.

Step 2: Factor Out the Common Factor

Factor \(x^{1/2}\) out of the equation: \[2x^{5/2} + 4x^{3/2} = 6x^{1/2} \Rightarrow x^{1/2}(2x^2 + 4x) = 6x^{1/2}\]

Step 3: Simplify the Equation

Since \(x^{1/2}\) is a common factor in every term, we can divide both sides of the equation by \(x^{1/2}\): \[2x^2 + 4x = 6\]

Step 4: Bring All Terms to One Side

Subtract 6 from both sides to set the equation to zero: \[2x^2 + 4x - 6 = 0\]

Step 5: Factor the Quadratic Equation

Now, let's factor the quadratic equation. Notice that 2 is also a common factor:\[2(x^2 + 2x - 3) = 0\]We factor \(x^2 + 2x - 3\) as:\[(x+3)(x-1) = 0\]This results in:\[2(x+3)(x-1) = 0\]

Step 6: Solve for x

Set each factor equal to zero and solve for \(x\):\[x+3 = 0 \Rightarrow x = -3\]\[x-1 = 0 \Rightarrow x = 1\] Thus, the solutions are \(x = -3\) and \(x = 1\).

Key Concepts

Common Factor

A common factor in an equation is a number or a variable that is present in all the terms of the equation. Identifying the common factor is the first crucial step in factoring any equation, as it simplifies the process immensely. In the equation \[ 2x^{5/2} + 4x^{3/2} = 6x^{1/2} \] each term shares the fractional power \(x^{1/2}\). This means we can factor out \(x^{1/2}\) from the entire equation.

• First, look at the smallest power of \(x\) in all terms.
• Factor this power out by dividing each term by it.

So, if each term has \(x^{1/2}\), you extract it, leaving simpler expressions in the parenthesis. This makes other algebra steps much more manageable.

Quadratic Equation

A quadratic equation is a type of polynomial that takes the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants. In our example, after simplifying and factoring out the common factor, we end up with: \[ 2x^2 + 4x - 6 = 0 \] This is a quadratic equation. Important characteristics of quadratic equations include:

• It always has an \(x^2\) term, making it a degree 2 polynomial.
• Quadratics can be factored further to find their roots.

Once in standard form, the next step is to factor, which often involves finding two binomials that multiply to produce the quadratic. Fields like algebra often rely on quadratic equations for a variety of applications, from physics to finance.

Fractional Power

Fractional powers may seem tricky, but they simplify expressions by allowing us to use simpler terms in equations. A fractional power represents a root, such as \(x^{1/2}\), which is equivalent to \(\sqrt{x}\). In our equation, notice how fractional powers were handled:

• Each term initially shared a base involving \(x\) with different fractional powers, like \(x^{5/2}\), \(x^{3/2}\).
• Identify and factor out the lowest fractional power shared among all terms, making the equation easier to solve.

Understanding fractional powers allows you to simplify equations and make substitutions that keep computations straightforward. It's a valuable tool for handling complex problems involving roots and radicals.

Solve for x

When we say "solve for \(x\)," it means finding the value or values of \(x\) that satisfy the equation. After factoring the quadratic equation \[ 2(x+3)(x-1) = 0 \] we need to set each factor equal to zero to solve for \(x\):

• Set \(x + 3 = 0\), solve to get \(x = -3\).
• Set \(x - 1 = 0\), solve to get \(x = 1\).

These solutions, \(x = -3\) and \(x = 1\), are the roots of the equation, meaning they make the whole expression equal to zero. Solving for \(x\) involves using algebraic manipulations and understanding the properties of equations and factors.