Algebra 1

Published by Prentice Hall
ISBN 10: 0133500403
ISBN 13: 978-0-13350-040-0

Chapter 9 - Quadratic Functions and Equations - 9-3 Solving Quadratic Equations - Lesson Check - Page 550: 1


$$x = \pm5$$

Work Step by Step

If you were to graph the equation you would find that the line intercepts the x-axis at $-5$ and $5$. To find this out from the equation only just solve for x. First step add 25 to both sides. You will get $$x^2 = 25$$ Next you would need to find the square root of 25 because that is the reverse of squaring something. To get the x to be alone, you must take off that squared symbol. And to do that you need to square root. And remember, whatever you do to one side of the equation you must do to the other so you make sure that the equation stays equal. So you find the square root of both sides. You will get $$x = \pm 5$$ Remember that the square root of a number is what number could be multiplied by itself to give you the number under the square root. It is a rule in math that any positive number under the square root (The number zero doesn't apply to this) yields you 2 answers. That would be the positive and the negative version of that number. That is because multiplying the positive version of the number by itself will give you the same answer as multiplying the negative number by itself. In this question, for example, the answer is $\pm5$ because $$5\times5 = 25$$ $$-5\times-5 = 25$$ $$So$$ $$\sqrt(25) = \pm5$$ The sign $\pm$ translates to Positive or Negative
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