Unformatted text preview: 10  x5  4 = 0 .
Solution
In this case we can reduce this to quadratic in form by using the substitution, u = x5 u 2 = x10 Using this substitution the equation becomes, 2u 2  u  4 = 0
This doesn’t factor and so we’ll need to use the quadratic formula on it. From the quadratic
formula the solutions are, u= 1 ± 33
4 Now, in order to get back to x’s we are going to need decimals values for these so, u= 1 + 33
= 1.68614
4 u= 1  33
= 1.18614
4 Now, using the substitution to get back to x’s gives the following, © 2007 Paul Dawkins 114 http://tutorial.math.lamar.edu/terms.aspx College Algebra 1 u = 1.68614 x5 = 1.68614 x = (1.68614 ) 5 = 1.11014 u = 1.18614 x5 = 1.18614 x = ( 1.18614 ) 5 = 1.03473 1 Of course we had to use a calculator to get the final answer for these. This is one of the reasons
that you don’t tend to see too many of these done in an Algebra class. The work and/or answers
tend to be a little messy. © 2007 Paul Dawkins 115 http://tutorial.math.lamar.edu/terms.aspx College Algebra Equations with Radicals
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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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