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Chapter 7: Problem 13

In Exercises \(5-14\) , determine whether the expression on the left of the equal sign is a difference of squares or a perfect square trinomial. If is, indicate which and then factor the expression and solve the equation for \(x\) . If the expression is in neither form, say so. $$ m^{2} x^{2}-n^{2}=0 $$

Short Answer

Expert verified
The expression is a difference of squares. Factored: \( (mx + n)(mx - n) \). Solutions for \( x \) are \( \pm \frac{n}{m} \).

Step by step solution

01

Recognize the Form of the Equation

Identify if the given expression on the left-hand side of the equation is a difference of squares or a perfect square trinomial. The given expression is \( m^{2} x^{2} - n^{2} = 0 \). Notice that \( m^{2} x^{2} \) and \( n^{2} \) are both perfect squares.
02

Write the Expression as a Difference of Squares

Since the given expression is in the form \( a^{2} - b^{2} \), it is a difference of squares where \( a = mx \) and \( b = n \).
03

Factor the Difference of Squares

Use the difference of squares formula: \( a^{2} - b^{2} = (a + b)(a - b) \). Thus, \( m^{2} x^{2} - n^{2} = (mx + n)(mx - n) \).
04

Solve the Equation for \( x \)

Set each factor equal to zero and solve for \( x \): 1. \( mx + n = 0 \)2. \( mx - n = 0 \)Solving both gives: 1. \( x = -\frac{n}{m} \)2. \( x = \frac{n}{m} \)

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Key Concepts

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

difference of squares
The difference of squares is a special algebraic expression that occurs when two perfect square terms are subtracted. The general form is \(a^2 - b^2\). This property reveals an elegant factorization: \(a^2 - b^2 = (a + b)(a - b)\). The key to recognizing a difference of squares is ensuring both terms are perfect squares.

If we look at the expression \(m^2 x^2 - n^2 = 0\), we see \(m^2 x^2\) and \(n^2\) are perfect squares. Consequently, \(m^2 x^2 - n^2\) can be factored using the difference of squares formula: \(a = mx\) and \(b = n\). So the equation becomes \(m^2 x^2 - n^2 = (mx + n)(mx - n)\). This simplification will help in solving the equation.
perfect square trinomial
A perfect square trinomial is another special algebraic expression that takes on the form \(a^2 + 2ab + b^2\) or \(a^2 - 2ab + b^2\). This type of expression simplifies to square of a binomial: either \((a + b)^2\) or \((a - b)^2\).

However, in our given problem, \(m^2 x^2 - n^2\) is a binomial, not a trinomial, and thus does not match the structure of a perfect square trinomial. Hence, we classify the equation as a difference of squares and proceed from there.
solving quadratic equations
Solving quadratic equations can be approached through different methods, one of the simplest being factoring. Given our equation \(m^2 x^2 - n^2 = 0\), after identifying it as a difference of squares, we factor it to obtain \( (mx + n)(mx - n) = 0\).

The next step is to set each factor to zero and solve for x.

  • \(mx + n = 0\)
  • \(mx - n = 0\)

Solving these separately:
For \(mx + n = 0\), subtract \(n\) and then divide by \(m\), getting \(x = -\frac{n}{m}\).
For \(mx - n = 0\), add \(n\) and then divide by \(m\), getting \(x = \frac{n}{m}\).

These solutions provide the values of \(x\) that satisfy the original equation.
factoring techniques
Factoring is a key technique in solving quadratic equations and more complex algebraic expressions. The fundamental idea is to express the equation in a product of simpler expressions that can be solved individually. There are several common factoring techniques:
  • Greatest Common Factor (GCF): Factor out the highest common factor from all terms.
  • Difference of Squares: For expressions of the form \(a^2 - b^2\).
  • Perfect Square Trinomial: For trinomial forms like \(a^2 + 2ab + b^2\) or \(a^2 - 2ab + b^2\).
  • Quadratic Form: For general trinomials \(ax^2 + bx + c\), factor into binomials.

In our example, identifying the expression as a difference of squares allowed us to use the appropriate factoring method to simplify and solve the equation efficiently. Understanding these techniques will empower you to tackle a variety of algebraic problems.

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Most popular questions from this chapter

History When the famous German mathematician Gauss was a young boy, he amazed his teacher by rapidly computing the sum of the integers from 1 to 100. He realized that he could compute the sum without adding all the numbers, by grouping the 100 numbers into pairs. To see a shortcut for finding this sum, look at two lists of 1 to 100, one in reverse order. \(\begin{array}{cccccccccccc}{1} & {2} & {3} & {4} & {5} & {6} & {7} & {\dots} & {50} & {\dots} & {94} & {95} & {96} & {97} & {98} & {99} & {100} \\ {100} & {99} & {98} & {97} & {96} & {95} & {94} & {\dots} & {51} & {\dots} & {7} & {6} & {5} & {4} & {3} & {2} & {1}\end{array}\) a. What is the sum of each pair? b. How many pairs are there? c. What is the sum of all these pairs? d. How many times is each of the integers from 1 to 100 counted in this sum? e. Consider your answers to Parts c and d. What is the sum of the integers from 1 to 100? f. Explain how you can use this same reasoning to find the sum of the integers from 1 to n for any value of n. Write a formula for s, the sum of the first n positive integers. g. Chloe added several consecutive numbers, starting at 1, and found a sum of 91. Write an equation you could use to find the numbers she added. Solve your equation by completing the square. Check your answer with the formula.

Sports Brianna and Lucita are playing tennis. On one volley, the height of the ball \(h,\) in feet, could have been described with the following equation, where \(t\) is the time in seconds since Brianna hit the ball: $$ h=-16(t-1)^{2}+20 $$ Assuming Lucita will let the ball bounce once, when will it hit the ground? Write and solve an equation to help you answer this question. Give your answer to the nearest hundreth of a second.

Solve each equation. $$ (r-8)^{2}+3=52 $$

Expand each expression. $$ (3 k-2 m)^{2} $$

Factor the expression on the left side of each equation as much as possible, and find all the possible solutions. It will help to remember that \(x^{4}=\left(x^{2}\right)^{2}, x^{8}=\left(x^{4}\right)^{2},\) and \(x^{3}=x\left(x^{2}\right) .\) $$ x^{4}-1=0 $$
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