/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 21 In Problems \(7-22,\) solve each... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 21

In Problems \(7-22,\) solve each inequality. 21\. \(6\left(x^{2}-1\right)>5 x\)

Short Answer

Expert verified
The solution is \( x \in (-\infty, -\frac{2}{3}) \cup (1.5, \infty) \).

Step by step solution

01

Expand the Expression

First, expand the left side of the inequality. Given the expression is \[ 6(x^2 - 1) > 5x \] Expanding gives: \[ 6x^2 - 6 > 5x \]
02

Reorganize the Inequality

Move all terms to one side of the inequality to set it to zero, so it can be solved as a quadratic inequality. This gives: \[ 6x^2 - 5x - 6 > 0 \]
03

Factor the Quadratic Expression

Factor the quadratic expression on the left-hand side. Set it to find the critical points where the expression equals zero:\[ 6x^2 - 5x - 6 = 0 \] Using the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \), with \( a = 6 \), \( b = -5 \), and \( c = -6 \), we get:\[ x = \frac{5 \pm \sqrt{25 + 144}}{12} \]Simplifying further, we get:\[ x = \frac{5 \pm 13}{12} \]Hence, the roots are \( x = 1.5 \) and \( x = -\frac{2}{3} \).
04

Test Intervals

Using the roots \( x = 1.5 \) and \( x = -\frac{2}{3} \), divide the number line into three intervals: \( (-\infty, -\frac{2}{3}) \), \( (-\frac{2}{3}, 1.5) \), and \( (1.5, \infty) \). Test a value from each interval in the inequality \[ 6x^2 - 5x - 6 > 0 \].
05

Validate Each Interval

For \( x = -1 \) (in the interval \( (-\infty, -\frac{2}{3}) \)): \[ 6(-1)^2 - 5(-1) - 6 = 6 + 5 - 6 = 5 > 0 \]For \( x = 0 \) (in the interval \( (-\frac{2}{3}, 1.5) \)): \[ 6(0)^2 - 5(0) - 6 = -6 < 0 \]For \( x = 2 \) (in the interval \( (1.5, \infty) \)): \[ 6(2)^2 - 5(2) - 6 = 24 - 10 - 6 = 8 > 0 \]So, the inequality holds for the intervals \( (-\infty, -\frac{2}{3}) \) and \( (1.5, \infty) \).
06

Write the Solution

Combine the intervals where the inequality holds. Therefore, the solution to the inequality is: \[ x \in (-\infty, -\frac{2}{3}) \cup (1.5, \infty) \]

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Key Concepts

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

Quadratic Inequality
Quadratic inequalities are expressions that involve a quadratic term (like \(x^2\)) and an inequality sign (>,<,鈮�,鈮�). To solve these inequalities, we need to find the range of values that satisfy the inequality. It鈥檚 different from solving quadratic equations because instead of finding exact values, we usually find intervals where the inequality holds true. For example, consider the inequality we have: \[6(x^2 - 1) > 5x.\] Our goal is to find all \(x\) values that make this inequality true.
Factoring
Factoring is a method used to break down quadratic expressions into simpler components called factors. These factors can be used to find the roots or solutions of the equation. For inequality problems, it's helpful to factor the quadratic expression to identify critical points or roots.
For our example, we rearranged the terms: \[6x^2 - 5x - 6 > 0\]
We then factorized it, but when factoring directly is complex, we can always use the quadratic formula. Either way, the factors reveal the critical points, which are \(x = 1.5\) and \(x = -\frac{2}{3}\).
Interval Testing
Interval testing is a strategy to determine where a quadratic inequality holds true by evaluating test points in different intervals formed by the roots. After finding the roots through factoring or the quadratic formula, we divide the number line into intervals.
  • For \(6x^2 - 5x - 6\), the roots \(x = 1.5\) and \(x = -\frac{2}{3}\) create three intervals.
  • We test points in these intervals to see where the inequality is satisfied.
For example:
In \((-\frac{2}{3}, 1.5)\), using \(x = 0\), we found \(6(0)^2 - 5(0) - 6 = -6\), which is less than 0, so this interval does not satisfy the inequality. But for intervals \((-鈭�, -\frac{2}{3})\) and \((1.5, 鈭�)\), chosen test points made the inequality true.
Quadratic Formula
The quadratic formula, \(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\), is used to find the roots of any quadratic equation. It's particularly useful when factoring is difficult or not straightforward.
For the inequality \[6x^2 - 5x - 6 = 0\], we identified the coefficients \(a = 6\), \(b = -5\), and \(c = -6\).
Plugging these into the quadratic formula, we got:
\[x = \frac{5 \pm 13}{12}\]
This simplification gave us the roots \(x = 1.5\) and \(x = -\frac{2}{3}\). These roots are essential for interval testing and determining where the inequality is satisfied.

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Most popular questions from this chapter

(a) find the vertex and the axis of symmetry of each quadratic function, and determine whether the graph is concave up or concave down. (b) Find the y-intercept and the \(x\) -intercepts, if any. (c) Use parts (a) and (b) to graph the function. (d) Find the domain and the range of the quadratic function. (e) Determine where the quadratic function is increasing and where it is decreasing. (f) Determine where \(f(x)>0\) and where \(f(x)<0\) \(f(x)=3 x^{2}+6 x+2\)

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The daily revenue \(R\) achieved by selling \(x\) boxes of candy is \(R(x)=9.5 x-0.04 x^{2}\). The daily cost \(C\) of selling \(x\) boxes of candy is \(C(x)=1.25 x+250 .\) (a) How many boxes of candy must the firm sell to maximize revenue? What is the maximum revenue? (b) Profit is given as \(P(x)=R(x)-C(x) .\) What is the profit function? (c) How many boxes of candy must the firm sell to maximize profit? What is the maximum profit? (d) Provide a reasonable explanation as to why the answers found in parts (a) and (c) differ. Explain why a quadratic function is a reasonable model for revenue.
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