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Chapter 0: Problem 78

Solve the equation by extracting square roots. $$ (x-5)^{2}=30 $$

Short Answer

Expert verified
The solutions for x are \(x = 5 + \sqrt{30}\) and \(x = 5 - \sqrt{30}\)

Step by step solution

01

Write down the equation

The given equation is \((x-5)^{2} = 30\).
02

Extract the Square Roots

Take the square root of both sides of the equation, noting that there should be two solutions (positive and negative square roots) for x. So, we write \(\sqrt{(x-5)^{2}} = \pm \sqrt{30}\) which gives \(x - 5 = \pm \sqrt{30}\).
03

Solve for x

Next, we solve for x by adding 5 to both sides of the equation: \(x = 5 \pm \sqrt{30}\).

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Key Concepts

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

Solving Quadratic Equations
Quadratic equations are a cornerstone in algebra with the general form of Ax^2 + Bx + C = 0. When tackling such problems, there are several methods available, including factoring, completing the square, using the quadratic formula, and extracting square roots. The choice of method often depends on the complexity of the equation and the coefficient values.

Solving quadratic equations by extracting square roots is applicable when the equation can be expressed as (x-h)^2 = k, where h and k are constants. This method involves isolating the squared term on one side and taking the square root of both sides, remembering to consider both the positive and negative roots. Understanding this approach is essential for handling equations that are perfect squares or can be easily manipulated into such a form.
Square Root Method
The square root method is a specialized technique for solving quadratic equations that simplifies the equation into the form (x-a)^2 = b. This method is ideal for equations that already contain a squared term. To use this method:
  • Isolate the squared term so that it is by itself on one side of the equation.
  • Take the square root of both sides, which yields two possible solutions since \( \sqrt{x^2} = \pm x \).
  • Solve for the variable by simplifying the roots and then isolating the variable on one side.
When applying the square root method, it's crucial to acknowledge both solutions that arise from the positive and negative square roots. This step is important since overlooking the negative solution can lead to an incomplete understanding of the problem's scope.
Radical Equations
Radical equations contain variables within a radical鈥攗sually a square root鈥攎aking them slightly more challenging than typical algebraic equations. To solve these equations, the following steps are commonly employed:
  • Isolate the radical on one side of the equation.
  • Square both sides of the equation to eliminate the radical, taking care to square the entire side.
  • Solve the resulting equation, which will often be a quadratic that may require further steps such as factoring or applying the quadratic formula.
  • Check your solutions in the original equation, as squaring both sides can sometimes introduce extraneous solutions that don't satisfy the original equation.
Understanding radical equations is key to mastering higher-level math concepts and ensuring that you consider all possible solutions within the correct domain.

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Most popular questions from this chapter

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

True or False? In Exercises 85 and 86, determine whether the statement is true or false. Justify your answer. If the inverse function of \(f\) exists and the graph of \(f\) has a \(y\)-intercept, the \(y\)-intercept of \(f\) is an \(x\)-intercept of \(f^{-1}\).

Find the average rate of change of the function from \(x_{1}\) to \(x_{2}\). $$ \begin{array}{cc} \text { Function } & x \text {-Values } \\ f(x)=3 x+8 \quad & x_{1}=0, x_{2}=3 \end{array} $$

In Exercises 55-68, determine whether the function has an inverse function. If it does, find the inverse function. $$ f(x)= \begin{cases}x+3, & x<0 \\ 6-x, & x \geq 0\end{cases} $$

In Exercises 39-54, (a) find the inverse function of \(f\), (b) graph both \(f\) and \(f^{-1}\) on the same set of coordinate axes, (c) describe the relationship between the graphs of \(f\) and \(f^{-1}\), and (d) state the domain and range of \(f\) and \(f^{-1}\). $$ f(x)=x^{3 / 5} $$
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