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Chapter 0: Problem 75

Solve the equation by extracting square roots. $$ (x-12)^{2}=16 $$

Short Answer

Expert verified
The solutions to the equation are x = 16 and x = 8.

Step by step solution

01

Isolate the square term

The first step is to isolate the square term, which in this case is already done for us. We have \((x - 12)^2 = 16\). This equation tells us that something (x-12), when squared, equals 16.
02

Apply square root on both sides

Take the square root of both sides of the equation. Remember that the square root of a number can be positive or negative. So, \(sqrt{(x - 12)^2} = sqrt{16}\) simplifies to \(x - 12 = 卤4\)
03

Solve for x

Solve the equation for x by adding 12 to both sides of the equation. Therefore, \(x = 12 卤 4\), which gives us two solutions: \(x = 16\) and \(x = 8\).

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Key Concepts

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

Extracting Square Roots
When working with quadratic equations, one of the methods to solve them is by extracting square roots. This method is particularly useful when the equation can be expressed in the form \[(x - a)^2 = b\]. Here, you have a perfect square on one side, which makes it possible to directly apply square roots to both sides of the equation. To apply this method effectively:
  • Check if the equation is already a perfect square. If not, you might need to manipulate it, although the example we're using is already simplified for demonstration.
  • Remember that extracting square roots introduces both positive and negative solutions, because squaring any real number always results in a positive value.
This becomes clearer when the equation transforms into two potential solutions. For example, \[(x - 12)^2 = 16 \] transforms to: \[x - 12 = 卤4\]. This indicates both possibilities: \[x - 12 = 4\] and \[x - 12 = -4\].
Isolation of Terms
The isolation of terms is a critical step in simplifying equations to make them solvable. In the context of quadratic equations, start by isolating the variable inside the square. This often involves rewriting the equation so that one side is simply the square of an expression and the other side is a constant.For example, you initially see: \[(x - 12)^2 = 16\]. Here, the square term \[(x - 12)^2\] stands alone on one side, making it ready for the next steps. If the equation were more complex, you might need to:
  • Add, subtract, multiply, or divide both sides of the equation to move other terms away from the squared term.
  • Ensure no other terms involve the square you are trying to isolate, so you can confidently apply the square root in the next step.
Isolation simplifies the equation and prepares it for solving.
Solving Equations
Once you've extracted square roots and isolated terms effectively, solving the equation becomes straightforward. From the equation \[(x - 12) = 卤4\], you solve for x by addressing both potential solutions:
  • First, handle the positive scenario: \[x - 12 = 4\], which simplifies to \[x = 16\].
  • Then, process the negative scenario: \[x - 12 = -4\], leading to \[x = 8\].
By returning to each fork in the road the positive and negative roots present, you uncover all valid solutions for x. This straightforward approach ensures that you account for both outcomes of squaring a number, as squaring inherently loses sign information. Thus, going back to solving ensures that no potential solutions are overlooked.

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Most popular questions from this chapter

(a) use the position equation \(s=-16 t^{2}+v_{0} t+s_{0}\) to write a function that represents the situation, (b) use a graphing utility to graph the function, (c) find the average rate of change of the function from \(t_{1}\) to \(t_{2}\), (d) interpret your answer to part (c) in the context of the problem, (e) find the equation of the secant line through \(t_{1}\) and \(t_{2}\), and (f) graph the secant line in the same viewing window as your position function. An object is thrown upward from ground level at a velocity of 96 feet per second. $$ t_{1}=2, t_{2}=5 $$

In Exercises 27 and 28, use the table of values for \(y=f(x)\) to complete a table for \(y=f^{-1}(x)\). $$ \begin{array}{|l|r|r|r|r|r|r|} \hline x & -2 & -1 & 0 & 1 & 2 & 3 \\ \hline f(x) & -2 & 0 & 2 & 4 & 6 & 8 \\ \hline \end{array} $$

Average Price The average prices \(p\) (in thousands of dollars) of a new mobile home in the United States from 1990 to 2002 (see figure) can be approximated by the model $$ p(t)= \begin{cases}0.182 t^{2}+0.57 t+27.3, & 0 \leq t \leq 7 \\ 2.50 t+21.3, & 8 \leq t \leq 12\end{cases} $$ where \(t\) represents the year, with \(t=0\) corresponding to 1990. Use this model to find the average price of a mobile home in each year from 1990 to 2002 . (Source: U.S. Census Bureau)

Find the average rate of change of the function from \(x_{1}\) to \(x_{2}\). $$ \begin{array}{cc} \text { Function } & x \text {-Values } \\ f(x)=x^{3}-3 x^{2}-x &\quad x_{1}=1, x_{2}=3 \end{array} $$

Find the average rate of change of the function from \(x_{1}\) to \(x_{2}\). $$ \begin{array}{cc} \text { Function } & x \text {-Values } \\ f(x)=-x^{3}+6 x^{2}+x &\quad x_{1}=1, x_{2}=6 \end{array} $$
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