document.write( "Question 32196: The section that I am on in my class is over contradiction. The equation is Show that neither x=1 or x=-1 can be solutions to the equation ax^2+bx+a=0 where a and b are integers with b odd. \n" ); document.write( "
Algebra.Com's Answer #18816 by mbarugel(146)\"\" \"About 
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\n" ); document.write( "Try plugging x = 1 into the equation. You get:\r
\n" ); document.write( "\n" ); document.write( "\"a%2A1+%2B+b%2A1+%2B+a+=0+\"
\n" ); document.write( "\"2a+%2B+b+=+0\"
\n" ); document.write( "\"b+=+-2a\"\r
\n" ); document.write( "\n" ); document.write( "However, b is odd, but 2a is even (two multiplied by any number yields an even number). Therefore, x=1 can't be a solution to the equation.\r
\n" ); document.write( "\n" ); document.write( "When plugging x = -1, we get:\r
\n" ); document.write( "\n" ); document.write( "\"a%2A1+%2Bb%2A%28-1%29+%2B+a+=+0\"
\n" ); document.write( "\"a+-+b+%2B+a+=+0\"
\n" ); document.write( "\"b+=+2a\"\r
\n" ); document.write( "\n" ); document.write( "Using the same reasoning, we conclude that x=-1 can't be a solution of your equation.\r
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\n" ); document.write( "\n" ); document.write( "I hope this helps!
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