Unformatted text preview: ial.math.lamar.edu/terms.aspx College Algebra So, what we’re going to need to do here is make sure that we’ve got a square root all by itself on
one side of the equation before squaring. Once that is done we can square both sides and the
square root really will disappear.
Here is the correct way to do this problem. y 4 = 4 y ( y4 ) 2 now square both sides = (4  y) 2 y  4 = 16  8 y + y 2
0 = y 2  9 y + 20
0 = ( y  5 )( y  4 ) Þ y = 4, y = 5 As with the first example we will need to make sure and check both of these solutions. Again,
make sure that you check in the original equation. Once we’ve square both sides we’ve changed
the problem and so checking there won’t do us any good. In fact checking there could well lead
us into trouble.
First y = 4 .
? 4+ 4 4 =4
4=4 OK So, that is a solution. Now y = 5 .
? 5+ 54 =4
? 5+ 1=4
6¹4 NOT OK So, as with the first example we worked there is in fact a single solution to the original equation,
y =4.
[Return to Problems] (b) 1 = t + 2t  3
Okay, so we will again need to get the square root on one side by itself before squ...
View
Full
Document
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

Click to edit the document details