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Chapter 7: Problem 38

Evaluate \(\sqrt{x-1}\) for \(x=17\)

Short Answer

Expert verified
The result for \(\sqrt{x-1}\) when \(x=17\) is 4.

Step by step solution

01

Substitute the given value

We have the equation \(\sqrt{x-1}\). We will include the value given which is \(x=17\), so the equation changes to \(\sqrt{17-1}\) or \(\sqrt{16}\).
02

Calculate the square root

Now that we have \(\sqrt{16}\), we just need to evaluate the square root. The square root of 16 is 4.

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Key Concepts

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

Square Root
The square root is a fundamental concept in mathematics. It helps in finding a number that, when multiplied by itself, gives the original number.
A square root is denoted by the radical symbol \( \sqrt{} \). For example, \( \sqrt{16} \) is the number that when multiplied by itself results in 16.
In simpler terms:
  • If \( a^2 = b \), then \( a \) is the square root of \( b \).
This exercise involves finding the square root of 16. Calculating this gives us 4, because \( 4 \times 4 = 16 \). This direct relationship helps simplify many mathematical problems, involving finding roots of numerical or algebraic expressions.
Substitution Method
The substitution method is a technique often used in mathematics to make solving equations easier. It involves placing a known value into an expression or equation to simplify it, as seen in this exercise.
For instance, in the given expression \( \sqrt{x-1} \), we are instructed to substitute \( x \) with 17.
Let's break down the steps:
  • Take the given value of \( x \), which is 17.
  • Replace \( x \) in the expression with 17, changing \( \sqrt{x-1} \) to \( \sqrt{17-1} \).
  • Simplify by subtracting: \( 17-1 = 16 \), resulting in \( \sqrt{16} \).
This method is invaluable in solving equations because it systematically reduces the number of variables until the expression is solved.
Algebra
Algebra is a branch of mathematics dealing with symbols and the rules for manipulating these symbols. It is the language through which we describe mathematical relationships.
The aim is to find the unknown or express it in terms of known quantities.
In our exercise, we're dealing with a simple algebraic expression \( \sqrt{x-1} \).
  • The expression involves a single variable \( x \), which we know the value of.
  • Algebra allows us to generalize mathematical statements and form equations that can be solved for one or more variables.
  • It's essential in problem-solving and is used across various fields, such as engineering, science, and economics.

Understanding algebra helps in evaluating expressions, finding solutions for equations, and modeling real-life situations effectively.

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Most popular questions from this chapter

denominators are opposites, or additive inverses. Add or subtract as indicated. Simplify the result, if possible. $$\frac{5 x-2}{3 x-4}+\frac{2 x-3}{4-3 x}$$

perform the indicated operation or operations. Simplify the result, if possible. $$\frac{6 b^{2}-10 b}{16 b^{2}-48 b+27}+\frac{7 b^{2}-20 b}{16 b^{2}-48 b+27}-\frac{6 b-3 b^{2}}{16 b^{2}-48 b+27}$$

denominators are opposites, or additive inverses. Add or subtract as indicated. Simplify the result, if possible. $$\frac{3-x}{x-7}-\frac{2 x-5}{7-x}$$

perform the indicated operation or operations. Simplify the result, if possible. $$\frac{b}{a c+a d-b c-b d}-\frac{a}{a c+a d-b c-b d}$$

denominators are opposites, or additive inverses. Add or subtract as indicated. Simplify the result, if possible. $$\frac{x-2}{x^{2}-25}-\frac{x-2}{25-x^{2}}$$
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