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Chapter 1: Problem 21

Solve each equation by factoring. [Hint for Exer cises 19-22: First factor out a fractional power.] $$ 2 x^{5 / 2}+4 x^{3 / 2}=6 x^{1 / 2} $$

Short Answer

Expert verified
\(x = 1\)

Step by step solution

01

Identify Common Factor

Look for the smallest power of \(x\) in each term. In this case, the smallest power is \(x^{1/2}\). We will factor out \(x^{1/2}\) from each term.
02

Factor Out \(x^{1/2}\)

Factor out \(x^{1/2}\) from each term in the equation:\[2x^{5/2} + 4x^{3/2} = 6x^{1/2}\]becomes:\[x^{1/2}(2x^2 + 4x) = 6x^{1/2}\]
03

Isolate Polynomial Equation

Since \(x^{1/2}\) is common in both sides, you can divide both sides by \(x^{1/2}\), assuming \(x eq 0\):\[2x^2 + 4x = 6\]
04

Bring Equation to Standard Form

Bring the equation to standard form for factoring:\[2x^2 + 4x - 6 = 0\]
05

Factor the Quadratic Equation

Factor the quadratic equation \(2x^2 + 4x - 6 = 0\):- Look for two numbers that multiply to \(-12\) (\(2 \times -6\)) and add to \(4\).- The numbers are \(6\) and \(-2\).- Rewrite and factor by grouping:\[2x^2 + 6x - 2x - 6 = 0\]\[2x(x + 3) - 2(x + 3) = 0\]\[(2x - 2)(x + 3) = 0\]
06

Solve for \(x\)

Using the zero product property, set each factor equal to zero and solve:- \(2x - 2 = 0\) gives \(x = 1\).- \(x + 3 = 0\) gives \(x = -3\).
07

Include the Domain Restriction from the Fractional Powers

Since the original equation involves fractional powers which require \(x \geq 0\), exclude any negative solutions: only \(x = 1\) is valid.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Factoring Techniques
Factoring is a crucial step in solving many algebraic equations. It enables us to simplify expressions and find solutions efficiently. When faced with an equation that involves fractional powers, such as exponential expressions of variables, it's helpful to recognize and factor out the smallest power. In this particular exercise, identifying a common factor, like \(x^{1/2}\), simplifies the equation significantly.

Factoring has many different techniques, including:
  • Finding the greatest common factor (GCF): This approach looks for the largest term shared across all components of an equation.
  • Factoring by grouping: When expressions can't be simplified directly, grouping common terms can reveal opportunities for factoring further.
  • Special formulas: Recognizing patterns such as perfect square trinomials or difference of squares can accelerate the factoring process.
To maximize the effectiveness of factoring, practice identifying these techniques in various exercises to build competence and speed.
Polynomial Equations
Polynomial equations consist of terms with variables raised to non-negative integer powers. They form the basis of algebraic expressions, providing the framework for both simple and complex mathematical problems. In our original exercise, after factoring out the smallest fractional power, we arrive at a polynomial equation \(2x^2 + 4x = 6\).

Solving polynomial equations typically follows these steps:
  • Move all terms to one side of the equation to create a standard form, which allows for straightforward factoring.
  • Factor the polynomial using appropriate techniques, like factoring by grouping or employing special factoring formulas.
  • Set each factor equal to zero, using the zero product property to solve for the variable.
Polynomial equations can have various solutions depending on their degree and structure. It's vital to analyze each problem carefully to apply the correct method.
Fractional Powers
Fractional powers, or exponents, represent roots of numbers or variables. They are integral in simplifying and solving equations involving complex terms that contain roots. The exercise initially contains terms with fractional powers: \(x^{5/2}\), \(x^{3/2}\), and \(x^{1/2}\). Understanding how to manipulate these powers is essential when approaching such problems.

Here are some key points about working with fractional powers:
  • \(x^{m/n}\) denotes the \(n\)-th root of \(x\) raised to the \(m\)-th power. For example, \(x^{1/2}\) is like saying \(\sqrt{x}\).
  • When factoring out a fractional power, consider it as the smallest common "root" that can simplify all terms in the equation.
  • Always be cautious of the domain of the variable, as negative bases can impose restrictions due to the roots.
In the given problem, acknowledging and using fractional powers helps in both simplification for ease of solving and for handling domain considerations.

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Most popular questions from this chapter

SOCIAL SCIENCE: Health Club Attendance A recent study analyzed how the number of visits a person makes to a health club varies with the monthly membership price. It found that the number of visits per year is given approximately by \(v(x)=-0.004 x^{2}+0.56 x+42\), where \(x\) is the monthly membership price. What monthly price maximizes the number of visits?

BUSINESS: Straight-Line Depreciation Straight-line depreciation is a method for estimating the value of an asset (such as a piece of machinery) as it loses value ( \({ }^{\prime \prime}\) depreciates" \()\) through use. Given the original price of an asset, its useful lifetime, and its scrap value (its value at the end of its useful lifetime), the value of the asset after \(t\) years is given by the formula: $$ \begin{aligned} \text { Value }=(\text { Price })-&\left(\frac{(\text { Price })-(\text { Scrap value })}{(\text { Useful lifetime })}\right) \cdot t \\ & \text { for } 0 \leq t \leq(\text { Useful lifetime }) \end{aligned} $$ a. A newspaper buys a printing press for $$\$ 800,000$$ and estimates its useful life to be 20 years, after which its scrap value will be $$\$ 60,000$$. Use the formula above Exercise 63 to find a formula for the value \(V\) of the press after \(t\) years, for \(0 \leq t \leq 20\) b. Use your formula to find the value of the press after 10 years. c. Graph the function found in part (a) on a graphing calculator on the window \([0,20]\) by \([0,800,000] .\) [Hint: Use \(x\) instead of \(t\).]

$$ \text { If } f(x)=a x, \text { then } f(f(x))=? $$

BUSINESS: Isocost Lines An isocost line (iso means "same") shows the different combinations of labor and capital (the value of factory buildings, machinery, and so on) a company may buy for the same total cost. An isocost line has equation $$ w L+r K=\mathrm{C} \quad \text { for } L \geq 0, \quad K \geq 0 $$ where \(L\) is the units of labor costing \(w\) dollars per unit, \(K\) is the units of capital purchased at \(r\) dollars per unit, and \(C\) is the total cost. Since both \(L\) and \(K\) must be nonnegative, an isocost line is a line segment in just the first quadrant. a. Write the equation of the isocost line with \(w=8, \quad r=6, \quad C=15,000\), and graph it in the first quadrant. b. Verify that the following \((L, K)\) pairs all have the same total cost. \((1875,0),(1200,900),(600,1700),(0,2500)\)

$$ \begin{array}{l} \text { For each function, find and simplify }\\\ \frac{f(x+h)-f(x)}{h} . \quad(\text { Assume } h \neq 0 .) \end{array} $$ $$ f(x)=2 x^{2}-5 x+1 $$
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