document.write( "Question 87746: Use the rational zeros theorem and the equation x4 – 12 = 0 to show that (12)¼ (i.e. the 4th root of 12) is irrational. \n" ); document.write( "
Algebra.Com's Answer #63694 by longjonsilver(2297)\"\" \"About 
You can put this solution on YOUR website!
i have never used this rational zeros theorem - i have had a quick look on the web and i think i understand, so here goes.\r
\n" ); document.write( "\n" ); document.write( "q is the factors of 1 --> the coefficient of the x^4 term
\n" ); document.write( "p is the factors of 12\r
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\n" ); document.write( "\n" ); document.write( "q = +1, -1
\n" ); document.write( "p = 1,2,3,4,6,12,-1,-2,-3,-4,-6,-12\r
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\n" ); document.write( "\n" ); document.write( "Now, N=p/q gives all RATIONAL values that when put into the polynomial and give zero are solutions. For our values of p and q, we get N=1,2,3,4,6,12,-1,-2,-3,-4,-6,-12 as possible solutions to the polynomial.\r
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\n" ); document.write( "\n" ); document.write( "However, putting each of these as x into the polynomial - none gives us zero when we raise it to the power 4 and then subtract 12.\r
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\n" ); document.write( "\n" ); document.write( "So, the conclusion is that if there is a real solution, it has to be irrational. Thinking about this particular case, we have \"+x%5E4=12+\" and so there is a real solution...there is a number that multiplies by itself 4 times to give 12 and it is between 1 and 2 since \"1%5E4+=+1\" and \"2%5E4=16\".\r
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\n" ); document.write( "\n" ); document.write( "cheers
\n" ); document.write( "Jon.
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