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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 \(x\) such that \(f(x)=0\) are \(x=2\sqrt{3}\) and \(x=-2\sqrt{3}\).

Step by step solution

01

Set the function equal to zero

To find the roots of the function, we need to set the function equal to zero and solve for \(x\). Thus, we have: \[0=\frac{12-x^{2}}{5}\]
02

Isolate \(x^{2}\)

Multiply both sides of the equation by 5 and bring \(x^{2}\) to one side: \[5*0=12-x^{2}\] Simplifying, we get: \[x^{2}=12\]
03

Solve for \(x\)

Once \(x^{2}=12\), use the principle of square roots, which states that if \(a^{2}=b\), then \(a=\sqrt{b}\) or \(a=-\sqrt{b}\). Thus, \[x=\sqrt{12}\] or \[x=-\sqrt{12}\] Since the square root of 12 simplifies to 2 times the square root of 3, the roots of the function are \(x=2\sqrt{3}\) and \(x=-2\sqrt{3}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Real Solutions
Real solutions are the values of the variable, such as \(x\), that satisfy the equation when it is set to zero. In the context of quadratic equations, real solutions mean that the roots are actual numbers and not imaginary. For example, when dealing with \(f(x) = 0\) in the equation \(f(x)=\frac{12-x^{2}}{5}\), we are searching for the real values of \(x\) that make this equation true.

Understanding real solutions is crucial because they represent the points where the function crosses the x-axis in a graph. In our exercise, we identify these solutions by solving the equation \(x^2 = 12\).
  • If the equation can be solved without using imaginary numbers, then the roots are real.
  • In general, when we calculate a square root to find our solutions鈥攚ithout needing any complex number setup鈥攚e confirm they are real solutions.
Square Roots
When solving quadratic equations, like \(x^2 = 12\), the square root is a fundamental tool. The square root function helps in determining the value of \(x\). Squaring a number means multiplying it by itself, so taking the square root is essentially asking "what number, when multiplied by itself, gives the original number?"

For example, since \(x^2 = 12\), we take square roots of both sides to find \(x\). The principle of square roots states that if \(a^2 = b\), then \(a = \sqrt{b}\) or \(a = -\sqrt{b}\). Therefore, \(x\) could be either \(\sqrt{12}\) or \(-\sqrt{12}\).
  • It's important to include both the positive and negative roots when taking square roots because both numbers will square to give the original value.
  • The square root of 12 simplifies to \(2\sqrt{3}\), so the solutions are \(x = 2\sqrt{3}\) and \(x = -2\sqrt{3}\).
Isolating Variables
Isolating the variable is often the first step to solve an equation. In our exercise, isolating \(x^2\) is crucial. This technique involves rearranging the equation to get the variable on one side of the equation to make solving straightforward. When you have \(\frac{12-x^{2}}{5} = 0\), isolating \(x^2\) will involve removing constants and any coefficients.

Here's the process step-by-step:
  • First, eliminate the fraction by multiplying both sides by 5 to get rid of the denominator: \(5 * 0 = 12 - x^2\), which simplifies to \(x^2 = 12\).
  • Now that \(x^2\) stands alone, we can easily take the square roots of both sides to find \(x\).
By isolating \(x^2\), we directly set ourselves up to use the square root principle. The clearer the equation is, the easier it is to find real solutions.

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Most popular questions from this chapter

The number \(N\) of bacteria in a refrigerated food is given by $$N(T)=10 T^{2}-20 T+600, \quad 2 \leq T \leq 20$$ where \(T\) is the temperature of the food in degrees Celsius. When the food is removed from refrigeration, the temperature of the food is given by $$T(t)=3 t+2, \quad 0 \leq t \leq 6$$ where \(t\) is the time in hours. (a) Find the composition \((N \circ T)(t)\) and interpret its meaning in context. (b) Find the bacteria count after 0.5 hour. (c) Find the time when the bacteria count reaches 1500 .

The frequency of vibrations of a piano string varies directly as the square root of the tension on the string and inversely as the length of the string. The middle A string has a frequency of 440 vibrations per second. Find the frequency of a string that has 1.25 times as much tension and is 1.2 times as long.

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Determine whether the function has an inverse function. If it does, then find the inverse function. $$f(x)=\left\\{\begin{array}{ll}-x, & x \leq 0 \\\x^{2}-3 x, & x>0\end{array}\right.$$

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