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Chapter 2: Problem 84

Suppose $$ 3 x^{2}+b x+7>0 $$ for every real number \(x\). Show that \(|b|<2 \sqrt{21}\).

Short Answer

Expert verified
To show that \(|b|<2 \sqrt{21}\), we analyzed the quadratic function \(f(x) = 3x^2 + bx + 7\) and its discriminant. We found that for the inequality to hold for all real numbers \(x\), the discriminant must be negative: \(b^2 - 84 < 0\), or \((b - \sqrt{84})(b + \sqrt{84}) < 0\). This inequality is satisfied when \(-\sqrt{84} < b < \sqrt{84}\), which implies that \(|b| < 2 \sqrt{21}\).

Step by step solution

01

Write down the quadratic function and analyze the discriminant

Given the quadratic function \(f(x) = 3x^2 + bx + 7\), we need to analyze the discriminant, which is given by \(\Delta = b^2 - 4ac\), where \(a = 3\), \(b\) is the coefficient we want to find, and \(c = 7\). To ensure that the parabola formed by \(f(x)\) never crosses the x-axis and stays above it, the discriminant should be negative, i.e., \(\Delta < 0\).
02

Compute the discriminant and find the inequality in terms of \(b\)

Compute the discriminant \(\Delta\) using the values of \(a\), \(b\), and \(c\): \[\begin{aligned} \Delta &= b^2 - 4ac \\ &= b^2 - 4(3)(7) \\ &= b^2 - 84 \end{aligned}\] Now, we need to find the inequality in terms of \(b\). Since we require \(\Delta < 0\), we have: \[b^2 - 84 < 0\]
03

Solve the inequality for \(b\)

Solving the quadratic inequality \(b^2 - 84 < 0\): \[\begin{aligned} b^2 &< 84 \\ b^2 - 84 &< 0 \\ (b - \sqrt{84})(b + \sqrt{84}) &< 0 \\ \end{aligned}\] This inequality holds when one of the factors is positive, and the other is negative. There are two possible cases: 1. \(b - \sqrt{84} < 0\) and \(b + \sqrt{84} > 0 \Rightarrow b < \sqrt{84}\) 2. \(b - \sqrt{84} > 0\) and \(b + \sqrt{84} < 0 \Rightarrow b\ > -\sqrt{84}\) Combining the two cases, we get that \(-\sqrt{84} < b < \sqrt{84}\).
04

Simplify the result and obtain the condition on \(b\)

Now, we simplify and express our result in terms of \(|b|\). Since \(\sqrt{84} = 2 \sqrt{21}\), we have: \[-2 \sqrt{21} < b < 2 \sqrt{21}\] which implies that \(|b| < 2 \sqrt{21}\).

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