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Unformatted text preview: elf on one side of the
equation then square both sides. At this point the process is different so we’ll see how to proceed
from this point once we reach it in the first example.
(a) 2 x  1  x  4 = 2
So, the first thing to do is get one of the square roots by itself. It doesn’t matter which one we get
by itself. We’ll end up the same solution(s) in the end. 2x 1 = 2 + x  4 ( 2x 1 ) = (2 +
2 x4 ) 2 2x 1 = 4 + 4 x  4 + x  4
2x 1 = 4 x  4 + x
Now, we still have a square root in the problem, but we have managed to eliminate one of them.
Not only that, but what we’ve got left here is identical to the examples we worked in the first part
of this section. Therefore, we will continue now work this problem as we did in the previous sets
of examples. ( x  1) 2 ( = 4 x4 ) 2 x 2  2 x + 1 = 16 ( x  4 )
x 2  2 x + 1 = 16 x  64
x 2  18 x + 65 = 0 ( x  13)( x  5 ) = 0 Þ x = 13, x = 5 Now, let’s check both possible solutions in the original equation. We’ll start with x = 13 2 (13)  1  13  4 = 2
? ? 25  9 = 2
53= 2
So...
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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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