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Chapter 4: Problem 110

Solve the following inequalities. (a) \(x+\frac{1}{x}<2\) (b) \(x \leqslant 1+\frac{2}{x}\) (c) \(\sqrt{x}

Short Answer

Expert verified
a) \(x > 0, x \neq 1\); b) \(-1 \leq x \leq 2\); c) \(x > 0\).

Step by step solution

01

Solve Part (a)

The inequality is \(x + \frac{1}{x} < 2\). Write it as \(x^2 - 2x + 1 > 0\) by multiplying both sides by \(x\) (assuming \(x > 0\)) and rearranging. It becomes \((x-1)^2 > 0\). This inequality holds for all \(x\) except where \(x = 1\). Thus, the solution is \(x > 0, x eq 1\).
02

Consider Part (b)

The inequality is \(x \leqslant 1+ \frac{2}{x}\). Rearrange to \(x - 1 \leqslant \frac{2}{x}\), and then multiply both sides by \(x\) (assuming \(x > 0\)) to get \(x^2 - x \leqslant 2\). Rearranging gives \(x^2 - x - 2 \leqslant 0\), which factors to \((x - 2)(x + 1) \leqslant 0\). Solve this inequality using a number line: the solution is \(-1 \leqslant x \leqslant 2\).
03

Solve Part (c)

The inequality is \(\sqrt{x} < x + \frac{1}{4}\). Square both sides to get \(x < (x + \frac{1}{4})^2\), which simplifies to \(x < x^2 + \frac{1}{2}x + \frac{1}{16}\). Simplify further to \(0 < x^2 + \frac{1}{2}x + \frac{1}{16} - x\), or \(x^2 - \frac{1}{2}x + \frac{1}{16} > 0\). Determine the roots of the quadratic \(x^2 - \frac{1}{2}x + \frac{1}{16}\) using the quadratic formula, which are complex. Hence, \(x^2 - \frac{1}{2}x + \frac{1}{16} > 0\) for all \(x\). But initially considering the domain, we need \(x > 0\) for the square root. Thus the solution is \(x > 0\).

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Key Concepts

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

Solving Quadratic Equations
Understanding how to solve quadratic equations is a crucial step in tackling many types of mathematical problems, including inequalities. Quadratic equations are in the form of \( ax^2 + bx + c = 0 \). These involve terms of degree two (the highest exponent is 2). The solutions to these equations are found using several methods. These methods include:
  • Factoring: Breaking down the equation into simpler binomial expressions.
  • Using the Quadratic Formula: Given by \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), which directly provides solutions.
  • Completing the Square: Rewriting the equation in the form \((x - p)^2 = q\), then solving for \(x\).
Knowing how to manipulate and solve these equations is vital when they appear as part of an inequality.
Inequality Solutions
Inequalities like \( x + \frac{1}{x} < 2 \) require techniques different from those of equations. They involve finding a range of solutions rather than specific values. Handling inequalities often means:
  • Rearranging terms: Get the inequality in a standard form.
  • Multiplying or dividing by negative numbers: Be careful, as reversing inequalities is necessary.
  • Checking domain restrictions: Consider where the expression is defined or undefined (like when dividing by zero).
For example, in Part (a), we rearrange to \((x-1)^2 > 0\), which is true for all \(x\) except \(x = 1\). Understanding the axis of symmetry and vertex of related quadratic expressions can also help visualize solution regions.
Mathematical Reasoning
Mathematical reasoning involves logical thought processes that allow one to solve problems methodically. Let鈥檚 consider how reasoning applied to inequalities helps in reaching the solution effectively:
  • Assumptions and Implications: Make logical assumptions to simplify expressions. Like assuming \(x > 0\) allows appealing multiplication in the problem.
  • Logical Deduction: If then statements drive stepwise solutions, e.g., deducing ranges or restrictions from rearranged inequalities.
  • Validation of Solutions: Always check if found solutions satisfy the original problem, ensuring no steps violate any mathematical rules.
Using reasoning in problem solving builds mathematical rigor and accuracy, as seen in confirming domain appropriateness for inequalities.
Number Line Analysis
Number line analysis is a visual tool used to determine which intervals satisfy an inequality. It helps us better understand the "where" and "why" of solutions:
  • Plotting Critical Points: Identify and plot points where the inequality might change direction, often where the quadratic expression equals zero.
  • Testing Intervals: Check different regions marked by critical points to see where the inequality holds true.
  • Understanding Behavior at Boundary Points: Determine whether these points are included in the solution set, indicated by open or closed circles on the line.
For instance, in solving \( (x - 2)(x + 1) \leq 0 \), plotting the roots \(x = -1, 2\) lets us test regions, concluding \(-1 \leq x \leq 2\). This visual representation ensures holistic comprehension of solution ranges.

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Most popular questions from this chapter

(a) Suppose you know the position vectors \(\mathbf{p}, \mathbf{q}, \mathbf{r}\) corresponding to the midpoints of the three sides of a triangle. Can you reconstruct the vectors \(\mathbf{x}, \mathbf{y}, \mathbf{z}\) corresponding to the three vertices? (b) Suppose you know the vectors \(\mathbf{p}, \mathbf{q}, \mathbf{r},\) s corresponding to the midpoints of the four sides of a quadrilateral. Can you reconstruct the vectors \(\mathbf{w}, \mathbf{x}\) \(\mathbf{y}, \mathbf{z}\) corresponding to the four vertices? (c) Suppose you know the vectors \(\mathbf{p}, \mathbf{q}, \mathbf{r}, \mathbf{s}, \mathbf{t}\) corresponding to the midpoints of the five sides of a pentagon. Can you reconstruct the vectors \(\mathbf{v}, \mathbf{w}, \mathbf{x},\) y, z corresponding to the five vertices? \(\triangle\) The previous five problems explore a common structural theme \(-\) namely the link between certain sums (or averages) and the original, possibly unknown, data. However this algebraic link was in every case embedded in some practical, or geometrical, context. The next few problems have been stripped of any context, leaving us free to focus on the underlying structure in a purely algebraic, or arithmetical, spirit.

Find the remainder when we divide \(x^{2013}+1\) by \(x^{2}+x+1\).

Let \(\alpha\) and \(\beta\) be the roots of the quadratic equation $$ x^{2}+b x+c=0 $$ (a)(i) Write \(\alpha^{2}+\beta^{2}\) in terms of \(b\) and \(c\) only. (ii) Write \(\alpha^{2} \beta+\beta^{2} \alpha\) in terms of \(b\) and \(c\) only. (iii) Write \(\alpha^{3}+\beta^{3}-3 \alpha \beta\) in terms of \(b\) and \(c\) only. (b)(i) Write \(\alpha-\beta\) in terms of \(b\) and \(c\) only. (ii) Write \(\alpha^{2} \beta-\beta^{2} \alpha\) in terms of \(b\) and \(c\) only. (iii) Write \(\alpha^{3}-\beta^{3}\) in terms of \(b\) and \(c\) only.

(a) Prove that the number \(\sqrt{2}+\sqrt{3}\) is irrational. (b) Prove that the number \(\sqrt{2}+\sqrt{3}+\sqrt{5}\) is irrational

Complex numbers \(a+b i\), where both \(a\) and \(b\) are integers, are called Gaussian integers. Try to formulate a version of the "division algorithm" for "division with remainder" (where the remainder is always "less than" the divisor in some sense) for pairs of Gaussian integers. Extend this to construct a version of the Euclidean algorithm to find the HCF of two given Gaussian integers. It is a profoundly erroneous truism ... that we should cultivate the habit of thinking what we are doing. The precise opposite is the case. Civilisation advances by extending the number of important operations which we can perform without thinking about them. Alfred North Whitehead \((1861-1947)\)
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