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Unformatted text preview: the integer and/or fractional solutions. In this case they are, t= 19 + 23 2
=
14
7 t= 19  23
= 3
14 Now, as with completing the square, the fact that we got integer and/or fractional solutions means
that we could have factored this quadratic equation as well.
[Return to Problems] (d) 3
1
= +1
y2 y So, an equation with fractions in it. The first step then is to identify the LCD. LCD : y ( y  2 )
So, it looks like we’ll need to make sure that neither y = 0 or y = 2 is in our answers so that we
don’t get division by zero.
Multiply both sides by the LCD and then put the result in standard form. æ 3 ö æ1 ö
÷ = ç + 1÷ ( y )( y  2 )
è y2ø è y ø ( y )( y  2 ) ç 3 y = y  2 + y ( y  2)
3y = y  2 + y2  2 y
0 = y2  4 y  2 Okay, it looks like we’ve got the following values for the quadratic formula. a =1 b = 4 c = 2 Plugging into the quadratic formula gives, © 2007 Paul Dawkins 100 http://tutorial.math.lamar.edu/terms.aspx College Algebra y=  ( 4 ) ±
( 4 )  4 (1)( 2 )
2 (1)
2 4 ± 24
2
4±2 6
=
2
= 2± 6
= Note that both of these are going to be solutions since neither of them are the values that we need
to avoid.
[Return to Problems] (e) 16 x  x 2 = 0
We saw an equation similar to this in the previous section when we were looking at factoring
equations and it would definitely be ea...
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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
 MrVinh

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