Answer
The equation has two solutions, $x=\pm k$.
Work Step by Step
We are given
$x^2-k^2=0$
$(x-k)(x+k)=0$
$x-k=0$ or $x+k=0$
$x=k$ or $x=-k$
Therefore, the equation has two solutions, $x=\pm k$.
Algebra 1: Common Core (15th Edition)
by Charles, Randall I.
Published by
Prentice Hall
ISBN 10:
0133281140
ISBN 13:
978-0-13328-114-9
Chapter 9 - Quadratic Functions and Equations - 9-4 Factoring to Solve Quadratic Equations - Practice and Problem-Solving Exercises - Page 571: 37
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