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Unformatted text preview: (c) 4 z 2 + 49 = 0 [Solution]
(d) ( 2t - 9 ) = 5 [Solution]
(e) ( 3 x + 10 ) + 81 = 0 [Solution]
There really isn’t all that much to these problems. In order to use the square root property all that
we need to do is get the squared quantity on the left side by itself with a coefficient of 1 and the
number on the other side. Once this is done we can use the square root property.
(a) x 2 - 100 = 0
This is a fairly simple problem so here is the work for this equation. x 2 = 100 x = ± 100 = ±10
So, there are two solutions to this equation, x = ±10 . Remember this means that there are really
© 2007 Paul Dawkins 90 http://tutorial.math.lamar.edu/terms.aspx College Algebra two solutions here, x = -10 and x = 10 .
[Return to Problems] (b) 25 y 2 - 3 = 0
Okay, the main difference between this one and the previous one is the 25 in front of the squared
term. The square root property wants a coefficient of one there. That’s easy enough to deal with
however; we’ll just divide both sides by 25. Here is the work for this equation. 25 y 2 = 3
y2 = 3
25 Þ 3
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This note was uploaded on 06/06/2012 for the course ICT 4 taught by Professor Mrvinh during the Spring '12 term at Hanoi University of Technology.
- Spring '12