document.write( "Question 353894: I have a hard exponent quadratic question.\r
\n" ); document.write( "\n" ); document.write( "5x^2/3 + 2x^4/3 -13 =0\r
\n" ); document.write( "\n" ); document.write( "My first inclination was to rearrange the terms to...\r
\n" ); document.write( "\n" ); document.write( "2x^4/3 + 5x^2/3 -13 =0\r
\n" ); document.write( "\n" ); document.write( "This puts it in the right order for substitution / y= x^2/3\r
\n" ); document.write( "\n" ); document.write( "Thus... 2y^2 + 5y -13 =0\r
\n" ); document.write( "\n" ); document.write( "This is where I'm stuck. I don't know how to unfold the answer with the unusual powers.\r
\n" ); document.write( "\n" ); document.write( "Thanks for your help.
\n" ); document.write( "Neil
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Algebra.Com's Answer #252981 by ewatrrr(24785)\"\" \"About 
You can put this solution on YOUR website!
HI,
\n" ); document.write( "2x^4/3 + 5x^2/3 -13 =0
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\n" ); document.write( "2x^4 + 5x^2 -39 =0 \r
\n" ); document.write( "\n" ); document.write( "*Note: This is a bi-quadratic form of an equation. In General:
\n" ); document.write( "The bi-quadratic equation is ax4 + bx2 + c = 0 and as it can be writen as:.
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\n" ); document.write( "\"a%28x2%29%5E2%2Bbx2%2Bc=0\" has the roots:
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\n" ); document.write( "\"x%5E2+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+\"
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\n" ); document.write( "\"x%5E2+=+%28-5+%2B-+sqrt%28+5%5E2-4%2A2%2A%28-39%29+%29%29%2F%282%2A2%29+\"
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\n" ); document.write( "\"x%5E2+=+%28-5+%2B-+sqrt%28+5%5E2-4%2A2%2A%28-39%29+%29%29%2F%282%2A2%29+\"
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\n" ); document.write( "\"x%5E2+=+%28-5+%2B-+sqrt%28+337%29%29%2F4+\"
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\n" ); document.write( "\"x+=+sqrt%28%28-5+%2B-+sqrt%28+337%29%29%2F4+%29\"
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\n" ); document.write( "\"x+=+%28-5+%2B+sqrt%28+337%29%29%2F4+\" or \"x+=+%28-5+-+sqrt%28+337%29%29%2F4+%29\"
\n" ); document.write( "or\r
\n" ); document.write( "\n" ); document.write( "\"x+=+-%28%28-5++%2B+sqrt%28+337%29%29%2F4%29+\" or \"x+=+-%28%28-5++-+sqrt%28+337%29%29%2F4%29+\"
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