Introductory Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-805-X
ISBN 13: 978-0-13417-805-9

Chapter 9 - Section 9.3 - The Quadratic Formula - Exercise Set - Page 648: 38


The solution set is {-3, 3}.

Work Step by Step

For this problem, we can factor out a $3$ to simplify it: $$3(x^2 - 9) = 0$$ Divide each side by $3$ to simplify the expression: $$x^2 - 9 = 0$$ We see that both the first and second terms are perfect squares, so we see that we can use the difference of two squares formula to factor. The difference of two squares can be factored as follows: $$A^2 - B^2 = (A + B)(A - B)$$ where $A$ is the square root of the first term and $B$ is the square root of the second term. For this particular problem, we take the square root of the first term $x^2$, which is $x$, and the square root of the second term $9$, which is $3$; these are $A$ and $B$, respectively, for our factoring needs. Let us substitute these two values into the formula: $$(x + 3)(x - 3) = 0$$ According to the zero product property, the product of factors is zero if at least one factor is zero. Therefore, we can set each factor to zero to solve for $x$: $$x + 3 = 0$$ We subtract $3$ from each side to solve for $x$: $$x = - 3$$ Let's look at the second factor: $$x - 3 = 0$$ We add $3$ to each side of the equation to solve for $x$: $$x = 3$$ The solution set is {-3, 3}.
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