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Chapter 2: Problem 26

Solve the inequality by factoring. $$12 x^{2}+5 x-2 \geq 0$$

Short Answer

Expert verified
The solution to the inequality is \(x \leq -1/4\) or \(x \geq 2/3\)

Step by step solution

01

Factor the Quadratic Expression

The quadratic inequality is \(12x^2 + 5x - 2 \geq 0\). To factor this, we'll look for two numbers that multiply to be \(a*c = -24\) and add to be \(b = 5\). Those two numbers are \(8\) and \(-3\). Using these numbers, we rewrite middle term to get: \(12x^2 + 8x - 3x - 2\). Factoring by grouping then gives us: \((4x + 1)(3x - 2) \geq 0\).
02

Find the Zeros

To find the zeros of the inequality, we set \(4x + 1 = 0\) and solve for \(x\) to get \(x = -1/4\). And then we set \(3x - 2 = 0\) and solve for \(x\) to get \(x = 2/3\).
03

Create and Test the Intervals

We take the interval \(-\infty < x < -1/4\), substitute a value (say -1) into the inequality to obtain: \(-1 < 0\), which is false. Hence \(-\infty < x < -1/4\) does not satisfy the inequality. Similarly, for the interval \(-1/4 < x < 2/3\), substitute 0 into the inequality to obtain: \(0 < 0\), which is false. Thus, \(-1/4 < x < 2/3\) does not satisfy the inequality. Lastly, for the interval \(x > 2/3\), substituting 1 into the inequality gives: \(8 > 0\), which is true. Thus, \(x > 2/3\) does satisfy the inequality.
04

Formulate the Solution

Combining all the results obtained, the solution to the inequality is \(x \leq -1/4\) or \(x \geq 2/3\)

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Key Concepts

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

Factoring Quadratics
To solve the inequality \(12x^2 + 5x - 2 \geq 0\), we begin with factoring the quadratic expression. Factoring is a technique used to simplify a quadratic equation, making it easier to solve or analyze. In this case, we need to express the quadratic in terms of two binomials. This process involves finding two numbers that multiply to \(a \cdot c = -24\) and also add to \(b = 5\).

Those numbers are 8 and -3. This allows us to break down the middle term of the polynomial, rewriting \(12x^2 + 5x - 2\) as \(12x^2 + 8x - 3x - 2\). Grouping these terms appropriately, we can then factor them as \((4x + 1)(3x - 2)\).

This step simplifies the inequality into a product of linear terms, making it easier to identify the critical points and test for solutions. Factoring not only simplifies solving inequalities like this one but also helps in sketching the graph of quadratic functions.
Quadratic Equations
Once the quadratic is factored into \((4x + 1)(3x - 2)\), we need to determine the zeros of the equation. Zeros, also known as roots or solutions, are values of \(x\) that make the equation equal to zero. These values will help us in our interval testing later.

To find the zeros, we set each factor equal to zero. For the factor \(4x + 1\), solving \(4x + 1 = 0\) gives us \(x = -1/4\). Similarly, solving \(3x - 2 = 0\) results in \(x = 2/3\). These zeros are critical points that divide the number line into intervals, which we will test to find the solution to the inequality.

Understanding quadratic equations and how to find their roots is fundamental in algebra, as it forms the basis for solving inequalities and other higher-order mathematical problems.
Interval Testing
With the roots \(x = -1/4\) and \(x = 2/3\), we divide the number line into intervals to determine where the inequality \((4x + 1)(3x - 2) \geq 0\) holds.

The intervals are:
  • \(-\infty < x < -1/4\)
  • \(-1/4 < x < 2/3\)
  • \(x > 2/3\)
We test each interval by choosing a test point in the interval and checking if it satisfies the inequality. For instance, in \(-\infty < x < -1/4\), we can test \(x = -1\). Substituting \(-1\) into the inequality gives a negative result, indicating this interval doesn't satisfy the inequality.

Repeat this process for the other intervals. In \(-1/4 < x < 2/3\), a test point like \(x = 0\) results in a false inequality, while for \(x > 2/3\), a positive result, with \(x = 1\), means this interval satisfies the inequality.

Interval testing is a powerful method to find where quadratic inequalities hold true. By breaking down complex expressions into testable regions, it provides a clearer view of the solution.

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Most popular questions from this chapter

When designing buildings, engineers must pay careful attention to how different factors affect the load a structure can bear. The following table gives the load in terms of the weight of concrete that can be borne when threaded rod anchors of various diameters are used to form joints. $$\begin{array}{cc} \text { Diameter (in.) } & \text { Load (1b) } \\ \hline 0.3750 & 2105 \\ 0.5000 & 3750 \\ 0.6250 & 5875 \\ 0.7500 & 8460 \\ 0.8750 & 11,500 \end{array}$$ (a) Examine the table and explain why the relationship between the diameter and the load is not linear. (b) The function $$f(x)=14,926 x^{2}+148 x-51$$ gives the load (in pounds of concrete) that can be borne when rod anchors of diameter \(x\) (in inches) are employed. Use this function to determine the load for an anchor with a diameter of 0.8 inch. (c) since the rods are drilled into the concrete, the manufacturer's specifications sheet gives the load in terms of the diameter of the drill bit. This diameter is always 0.125 inch larger than the diameter of the anchor. Write the function in part (b) in terms of the diameter of the drill bit. The loads for the drill bits will be the same as the loads for the corresponding anchors. (Hint: Examine the table of values and see if you can present the table in terms of the diameter of the drill bit.)

In Exercises \(101-104,\) let \(f(t)=3 t+1\) and \(g(x)=x^{2}+4\). Evaluate \((f \circ f)(2)\)

The height of a ball after being dropped from the roof of a 200 -foot-tall building is given by \(h(t)=-16 t^{2}+200,\) where \(t\) is the time in seconds since the ball was dropped, and \(h(t)\) is in feet. (a) When will the ball be 100 feet above the ground? (b) When will the ball reach the ground? (c) For what values of \(t\) does this problem make sense (from a physical standpoint)?

In Exercises \(97-100,\) let \(f(t)=-t^{2}\) and \(g(x)=x^{2}-1\). Find an expression for \((g \circ g)(x),\) and give the domain of \(g \circ g\).

In Exercises \(101-104,\) let \(f(t)=3 t+1\) and \(g(x)=x^{2}+4\). Find an expression for \((f \circ f)(t)\), and give the domain of \(f \circ f\)
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