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Chapter 2: Problem 14

Solve the inequalities. Suggestion: A calculator may be useful for approximating key numbers. $$3 x^{2}+x \leq 4$$

Short Answer

Expert verified
The solution is \(-\frac{4}{3} \leq x \leq 1\).

Step by step solution

01

Rearrange the Inequality

First, rearrange the inequality to get everything on one side of the inequality sign. Subtract 4 from both sides:\[ 3x^2 + x - 4 \ ext{Now the inequality is:} \ 3x^2 + x - 4 \leq 0 \]
02

Find Critical Points

Find the roots of the equation by setting \( 3x^2 + x - 4 = 0 \). Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 3 \), \( b = 1 \), and \( c = -4 \):\[ x = \frac{-1 \pm \sqrt{1^2 - 4(3)(-4)}}{2(3)} \]Simplify inside the square root:\[ x = \frac{-1 \pm \sqrt{1 + 48}}{6} = \frac{-1 \pm \sqrt{49}}{6} \]So the roots are:\[ x = \frac{-1 + 7}{6} = 1 \ x = \frac{-1 - 7}{6} = -\frac{4}{3} \]
03

Test Intervals Between Critical Points

Test the intervals determined by the roots, which are \( x < -\frac{4}{3} \), \( -\frac{4}{3} < x < 1 \), and \( x > 1 \). Choose test points in each interval and substitute back into the inequality \( 3x^2 + x - 4 \leq 0 \).1. **Test point **\( x = -2 \) (Interval: \( x < -\frac{4}{3} \)): \[ 3(-2)^2 + (-2) - 4 = 12 - 2 - 4 = 6 \] (not \( \leq 0 \))2. **Test point** \( x = 0 \) (Interval: \( -\frac{4}{3} < x < 1 \)): \[ 3(0)^2 + 0 - 4 = -4 \] (\( \leq 0 \))3. **Test point** \( x = 2 \) (Interval: \( x > 1 \)): \[ 3(2)^2 + 2 - 4 = 12 + 2 - 4 = 10 \] (not \( \leq 0 \))
04

Determine Solution Set

From the test intervals, the inequality \( 3x^2 + x - 4 \leq 0 \) holds for the interval \( -\frac{4}{3} \leq x \leq 1 \). Include \( -\frac{4}{3} \) and \( 1 \) because they satisfy the inequality when substituted into \( 3x^2 + x - 4 = 0 \).

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Key Concepts

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

Quadratic Formula
The quadratic formula is a powerful tool used to find the roots of any quadratic equation in the form \( ax^2 + bx + c = 0 \). When we speak about the roots, these are simply the values of \( x \) which make the equation equal to zero.
The quadratic formula is given by:
  • \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
Here, \( a \), \( b \), and \( c \) are the coefficients of the quadratic equation, corresponding to the terms \( ax^2 \), \( bx \), and \( c \).
These coefficients help in calculating the
  • Discriminant \( b^2 - 4ac \), which determines the nature of the roots.
If the discriminant is positive, the equation has two real roots. If it is zero, there's precisely one real root. And if it's negative, the roots are complex numbers.
Inequalities
Inequalities are mathematical expressions that involve relationships between quantities. They show how one expression is larger or smaller compared to another. In the quadratic inequality \( 3x^2 + x - 4 \leq 0 \), our goal is to determine the range of \( x \) that makes this inequality true.
We can start solving such problems by finding the roots of the related quadratic equation. These roots act as critical points, dividing the number line into different intervals.
To decide which intervals satisfy the inequality, we test points from each segment and evaluate them in the original inequality.
  • If the result of the expression is \( \leq 0 \), the interval is part of our solution set.
  • This technique ensures we account for all possibilities and correctly identify all parts of the solution.
Roots of Equations
Roots of equations are the values of the variable that solve the equation. For a quadratic equation such as \( 3x^2 + x - 4 = 0 \), the roots represent the points where the function intersects the x-axis.
By using the quadratic formula, we calculated the roots \( x = 1 \) and \( x = -\frac{4}{3} \). These roots are critical because they not only help solve the equation but also have significant implications for inequalities.
Understanding the roots allows us to deduce the behavior of the corresponding quadratic function. Between these roots, if the quadratic opens upwards (as in this example), the function is negative, indicating the interval where the inequality holds.
  • The critical points (roots) are included in the solution set because the inequality is "less than or equal to zero".
  • This understanding is fundamental in solving inequalities as it shows you the exact range of \( x \) values that satisfy the given inequality expression.

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Most popular questions from this chapter

Find all real solutions of each equation. For Exercises \(31-36,\) give two forms for each answer: an exact answer (involving a radical) and a calculator approximation rounded to two decimal places. $$x^{-4}+x^{-2}+1=0$$

Find all real solutions of each equation. For Exercises \(31-36,\) give two forms for each answer: an exact answer (involving a radical) and a calculator approximation rounded to two decimal places. $$9 x^{4 / 3}-10 x^{2 / 3}+1=0$$

Determine all of the real-number solutions for each equation. (Remember to check for extraneous solutions.) $$\sqrt{x}+6=x$$

Use the discriminant to determine how many real roots each equation has. $$4 x^{2}-28 x+49=0$$

Let \(a, b,\) and \(c\) be nonnegative numbers. Follow steps (a) through (e) to show that $$\sqrt[3]{a b c} \leq \frac{a+b+c}{3}$$ with equality holding if and only if \(a=b=c\). This result is known as the arithmetic-geometric mean inequality for three numbers. (Applications are developed in the projects at the ends of Sections 4.6 and \(4.7 .\) ) (a) By multiplying out the right-hand side, show that the following equation holds for all real numbers \(A, B\) and \(C\) $$\begin{aligned} 3 A B C=& A^{3}+B^{3}+C^{3}-\frac{1}{2}(A+B+C) \\ & \times\left[(A-B)^{2}+(B-C)^{2}+(C-A)^{2}\right] \quad \quad (1) \end{aligned}$$ (b) Now assume for the remainder of this exercise that \(A\) \(B,\) and \(C\) are nonnegative numbers. Use equation (1) to explain why $$3 A B C \leq A^{3}+B^{3}+C^{3}\quad \quad (2)$$ (c) Make the following substitutions in inequality ( 2 ): \(A^{3}=a, B^{3}=b,\) and \(C^{3}=c .\) Show that the result can be written $$\sqrt[3]{a b c} \leq \frac{a+b+c}{3} \quad \quad (3)$$ (d) Assuming that \(a=b=c,\) show that inequality (3) becomes an equality. (e) Assuming \(\sqrt[3]{a b c}=\frac{a+b+c}{3},\) show that \(a=b=c\) Hint: In terms of \(A, B,\) and \(C,\) the assumption becomes \(A B C=\frac{A^{3}+B^{3}+C^{3}}{3} .\) Use this to substitute for \(A B C\) on the left-hand side of equation (1). Then use the resulting equation to deduce that \(A=B=C,\) and consequently \(a=b=c\)
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