Chapter 3.5, Problem 95PS

### Intermediate Algebra

10th Edition
Jerome E. Kaufmann + 1 other
ISBN: 9781285195728

Chapter
Section

### Intermediate Algebra

10th Edition
Jerome E. Kaufmann + 1 other
ISBN: 9781285195728
Textbook Problem

# Consider the following solution: 6 x 2 − 24 = 0 6 ( x 2 − 4 ) = 0 6 ( x + 2 ) ( x − 2 ) = 0 6 = 0 or x + 2 = 0 or x − 2 = 0 6 = 0 or x = − 2 or x = 2 The solution set is { − 2 , 2 } .Is this a correct solution? Would you have any suggestion to offer the person who used this approach?

To determine

To find:

The given solution correct or incorrect and offer the suggestion to the person who used this approach.

Explanation

Approach:

i) The Difference of Two Squares pattern is a2âˆ’b2=(a+b)(aâˆ’b).

ii) Multiplication is commutative, so the order of writing the factors is not important. For example, (a+b)(aâˆ’b) can also be written as (aâˆ’b)(a+b).

iii) abc=0 if and only if a=0 or b=0 or c=0.

The following steps have been followed to find out the solutions:

Step-1: Find out the common monomial factor and divide out from the equation.

Step-2: Convert the equation from the step-1 in the form of 2 squares with subtraction and compare with the difference of two squares pattern.

Step-3: Apply the difference of two squares pattern to the equation available in the step-2.

Step-4: Solve the equation and write the solutions set.

Calculation:

1) The given equation is 6x2âˆ’24=0.

The common monomial factor available in the above equation is 6.

So, the equation can be written as, 6(x2âˆ’4)=0.

Divide both sides of the equation by 6. Remember that this process has been skipped in the approach.

Therefore, x2âˆ’4=0.

2) The above equation can be written as,

x2âˆ’22=0.

Because, 22=2â‹…2=4

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