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Chapter 1: Problem 42

Find all real values of \(x\) such that \(f(x)=0\). $$f(x)=x^{2}-8 x+15$$

Short Answer

Expert verified
The solutions are \(x = 3\) and \(x = 5\).

Step by step solution

01

Set the function equal to zero

First, substitute \(f(x)\) with zero to form the following equation: \(0 = x^{2} - 8x + 15\).
02

Factor the quadratic equation

The equation \(0 = x^{2} - 8x + 15\) is a quadratic equation that can be factored. The factors of 15 that add up to -8 are -3 and -5. Therefore, \(x^{2} - 8x + 15\) can be factored into \((x-3)(x-5)\)
03

Set each factor equal to zero

The zero property of multiplication states that if \(a*b = 0\), then either \(a = 0\) or \(b = 0\). Therefore, setting each of the factors equal to zero results in two equations: \(x-3 = 0\) and \(x-5 = 0\).
04

Solve for x

Finally, solving each factor equal to zero results in two possible solutions for \(x\): \(x=3\) when \(x-3 = 0\) and \(x=5\) when \(x-5 = 0\). These are the two real solutions that make \(f(x)=0\)

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Most popular questions from this chapter

Use a graphing utility to graph the function. Be sure to choose an appropriate viewing window. $$f(x)=(x-1)^{3}+2$$

Sketch the graph of the function. $$f(x)=\left\\{\begin{array}{ll}1-(x-1)^{2}, & x \leq 2 \\\\\sqrt{x-2}, & x>2\end{array}\right.$$

Sketch the graph of the function. $$f(x)=\left\\{\begin{array}{ll}\sqrt{4+x}, & x<0 \\\\\sqrt{4-x}, & x \geq 0\end{array}\right.$$

Use a graphing utility to graph the function. Be sure to choose an appropriate viewing window. $$g(x)=|x|-5$$

Sketch the graph of the function. $$g(x)=[[x]]-1$$
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