Algebra 2 Common Core

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

Chapter 4 - Quadratic Functions and Equations - Chapter Test - Page 273: 14

Answer

The solutions to this equation are $x = 5, -5$.

Work Step by Step

We can solve this quadratic equation by factoring. We see that this equation can be factored using the formula to factor the difference of two squares, which is: $$a^2 - b^2 = (a + b)(a - b)$$ In the equation $x^2 - 25 = 0$, $a$ is the $\sqrt {x^2}$ or $x$, and $b$ is $\sqrt {25}$ or $5$. We plug these values into the formula: $$(x + 5)(x - 5) = 0$$ 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's set the first factor equal to $0$: First factor: $x + 5 = 0$ Subtract $5$ from each side to solve for $x$: $x = -5$ Second factor: $x - 5 = 0$ Add $5$ to each side of the equation to solve for $x$: $x = 5$ The solutions to this equation are $x = 5, -5$.
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