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Chapter 2: Problem 18

Solve the inequality. Then graph the solution set. $$x^{2}-6 x+9<16$$

Short Answer

Expert verified
The solution to the inequality \(x^{2} - 6x + 9<16\) is \(x\) belongs to (-1, 7), with 'belongs to' indicating that it includes all the values within this open interval.

Step by step solution

01

Set the inequality as an equation

To solve for \(x\), first set the inequality as an equation: \(x^{2}-6x+9=16\)
02

Simplify the equation and solve for \(x\)

Then, simplify the equation to \(x^{2} - 6x - 7 = 0\). This is a quadratic equation in the form \(ax^{2} + bx + c = 0\), so solve for \(x\) by factoring or using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Applying the quadratic formula, the roots are \(x = 7\) and \(x = -1\)
03

Identify Intervals and Determine the Solution Set

Now that \(x\) values are found, the real number line is split into three intervals: (-\infty, -1), (-1, 7) and (7, \infty). As for whether to use an open or closed circle at -1 and 7, when looking back at the original inequality, since it's '<' not '<=', open circles are used. Now, test a number from each interval in the original inequality. For example, take -2, 0, and 8 for each interval respectively. So, after testing, the inequality is true for the intervals (-1, 7), thus, the solution set is (-1, 7).

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Most popular questions from this chapter

Explore transformations of the form \(g(x)=a(x-h)^{5}+k\) (a) Use a graphing utility to graph the functions \(y_{1}=-\frac{1}{3}(x-2)^{5}+1\) and \(y_{2}=\frac{3}{5}(x+2)^{5}-3\) Determine whether the graphs are increasing or decreasing. Explain. (b) Will the graph of \(g\) always be increasing or decreasing? If so, then is this behavior determined by \(a, h,\) or \(k ?\) Explain. (c) Use the graphing utility to graph the function \(H(x)=x^{5}-3 x^{3}+2 x+1\) Use the graph and the result of part (b) to determine whether \(H\) can be written in the form \(H(x)=a(x-h)^{5}+k\) Explain.

Determine (if possible) the zeros of the function \(g\) when the function \(f\) has zeros at \(x=r_{1}, x=r_{2},\) and \(x=r_{3}\) $$g(x)=f(-x)$$

Use Descartes's Rule of Signs to determine the possible numbers of positive and negative real zeros of the function. $$h(x)=2 x^{4}-3 x+2$$

Use the position equation $$s=-16 t^{2}+v_{0} t+s_{0}$$ where \(s\) represents the height of an object (in feet), \(v_{0}\) represents the initial velocity of the object (in feet per second), \(s_{0}\) represents the initial height of the object (in feet), and \(t\) represents the time (in seconds). A projectile is fired straight upward from ground level \(\left(s_{0}=0\right)\) with an initial velocity of 160 feet per second. (a) At what instant will it be back at ground level? (b) When will the height exceed 384 feet?

Use the given zero to find all the zeros of the function. Function \(g(x)=x^{3}-7 x^{2}-x+87\) Zero \(5+2 i\)
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