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Chapter 2: Problem 82

Determine whether the statement is true or false. Justify your answer. The solution set of the inequality \(\frac{3}{2} x^{2}+3 x+6 \geq 0\) is the entire set of real numbers.

Short Answer

Expert verified
The statement is true. The solution set of the inequality \(\frac{3}{2} x^{2} + 3x + 6 \geq 0\) is indeed the entire set of real numbers because its graph is always above or on the x-axis for all real numbers.

Step by step solution

01

Establish the quadratic

The inequality provided is \(\frac{3}{2} x^{2} + 3x + 6 \geq 0\). This is clearly a quadratic inequality in the standard form \(ax^2 + bx + c\).
02

Compute the Discriminant

The Discriminant of a quadratic equation is given by \(D = b^2 - 4ac\). Here, \(a = \frac{3}{2}\), \(b = 3\) and \(c = 6\). Substituting these values into the Discriminant formula, we find that \(D = 3^2 - 4 * \frac{3}{2} * 6\) which simplifies to \(D = -9\).
03

Analyze the Discriminant

Since the Discriminant is less than zero, this means that the quadratic equation has no real roots. That is, there are no values of \(x\) for which \( \frac{3}{2} x^{2} + 3x + 6 = 0\).
04

Analyze the Coefficients

Since the coefficient of \(x^2\) is greater than zero, the parabola defined by the inequality opens upwards. And since there are no real roots, the parabola never crosses or touches the x-axis.
05

Finalize the conclusion

Given that the parabola opens upwards and never crosses the x-axis, the graph of the function is always above the x-axis. This implies the function is always greater than or equal to zero. Therefore, the solution to the inequality is the entire set of real numbers.

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Most popular questions from this chapter

Write a rational function \(f\) that has the specified characteristics. (There are many correct answers.) (a) Vertical asymptote: \(x=2\) Horizontal asymptote: \(y=0\) Zero: \(x=1\) (b) Vertical asymptote: \(x=-1\) Horizontal asymptote: \(y=0\) Zero: \(x=2\) (c) Vertical asymptotes: \(x=-2, x=1\) Horizontal asymptote: \(y=2\) Zeros: \(x=3, x=-3\), (d) Vertical asymptotes: \(x=-1, x=2\) Horizontal asymptote: \(y=-2\) Zeros: \(x=-2, x=3\)

Solve the inequality and graph the solution on the real number line. \(\frac{x^{2}-1}{x}<0\)

Determine whether each value of \(x\) is a solution of the inequality. Inequality. \(x^{2}-3<0 \quad\) (a) \(x=3 \quad\) (b) \(x=0\) (c) \(x=\frac{3}{2}\) (d) \(x=-5\)

Find the key numbers of the expression. \(\frac{x}{x+2}-\frac{2}{x-1}\)

Determine whether the statement is true or false. Justify your answer. The zeros of the polynomial \(x^{3}-2 x^{2}-11 x+12 \geq 0\) divide the real number line into four test intervals.
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