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Chapter 1: Problem 40

Find all real values of \(x\) such that \(f(x)=0\). $$f(x)=\frac{12-x^{2}}{5}$$

Short Answer

Expert verified
The real values of \(x\) that make this function zero are \(x = 2\sqrt{3}\) and \(x = -2\sqrt{3}\).

Step by step solution

01

Set the function equal to zero

To find the zeros of the function \(f(x)\), set \(f(x)\) equal to zero. In other words, the equation to solve is: \[ \frac{12-x^{2}}{5} = 0\]
02

Simplify the equation

Multiplying both sides by 5 (to get rid of the denominator), we get:\[ 12-x^{2} = 0.\] The equation simplifies further, after moving \(x^{2}\) to the other side, to: \[x^{2} = 12.\]
03

Solve for x

The next step is to take the square root of both sides to solve for \(x\). It's important to remember that when taking the square root of a value, the result is positive and negative. Hence, we get the solutions: \[x = \sqrt{12}\] and \[x = -\sqrt{12},\] which simplify further to \(x = 2\sqrt{3}\) and \(x = -2\sqrt{3}\)

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