Algebra 2 Common Core

Published by Prentice Hall
ISBN 10: 0133186024
ISBN 13: 978-0-13318-602-4

Chapter 5 - Polynomials and Polynomial Functions - 5-3 Solving Polynomial Equations - Practice and Problem-Solving Exercises - Page 302: 69

Answer

The solutions are $x = 6, \text{ and } x=-6$.

Work Step by Step

Rewrite the equation in standard form: $x^2 - 36 = 0$ We see that this equation can be factored using the formula to factor the difference of two squares. According to this formula: $a^2 - b^2 = (a + b)(a - b)$ In the equation $x^2 - 36 = 0$, $a$ is the $\sqrt {x^2}$ or $x$, and $b$ is $\sqrt {36}$ or $6$. We plug these values into the formula: $(x + 6)(x - 6) = 0$ The Zero-Product Property states that if the product of two factors equals $0$, then either factor equals $0$, or both factors equal $0$. Therefore, we can set each factor equal to $0$ to solve for $x$. Let us set the first factor equal to $0$: $x + 6 = 0$ $x = -6$ Let us set the second factor equal to $0$: $x - 6 = 0$ $x = 6$ The solutions are $x = -6, 6$. To check if the solution is correct, we plug in the values we got for $x$ into the original equation to see if both sides are equal. Let's plug in $6$ first: $(6)^2 + 1 = 37$ $36 + 1 = 37$ $37 = 37$ This solution is correct. Let us look at $-6$: $(-6)^2 + 1 = 37$ $36 + 1 = 37$ $37 = 37$ This solution is also correct.
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