Algebra: A Combined Approach (4th Edition)

Published by Pearson
ISBN 10: 0321726391
ISBN 13: 978-0-32172-639-1

Chapter 6 - Section 6.6 - Solving Quadratic Equations by Factoring - Exercise Set: 58

Answer

The solutions are 0 and 7.

Work Step by Step

$x^{3}$-14$x^{2}$+49x=0 x($x^{2}$-14$x^{}$+49)=0 x($x^{2}$-7$x^{}$-7x+49)=0 x[x(x-7)-7(x-7)]=0 x(x-7)(x-7)=0 x=0 or x-7=0 or x-7=0 x=0 or x=7 or x=7 The solutions are 0 and 7. Check Let x=0 $x^{3}$-14$x^{2}$+49x=0 $0^{3}$-14*$0^{2}$+49*0=0 0-0+0=0 0=0 Let x=7 $x^{3}$-14$x^{2}$+49x=0 $7^{3}$-14*$7^{2}$+49*7=0 343-14*49+343=0 343-686+343=0 686-686=0 0=0
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