document.write( "Question 353894: I have a hard exponent quadratic question.\r
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document.write( "5x^2/3 + 2x^4/3 -13 =0\r
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document.write( "My first inclination was to rearrange the terms to...\r
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document.write( "2x^4/3 + 5x^2/3 -13 =0\r
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document.write( "This puts it in the right order for substitution / y= x^2/3\r
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document.write( "Thus... 2y^2 + 5y -13 =0\r
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document.write( "This is where I'm stuck. I don't know how to unfold the answer with the unusual powers.\r
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document.write( "Thanks for your help.
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document.write( "Neil \n" );
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Algebra.Com's Answer #252981 by ewatrrr(24785)![]() ![]() You can put this solution on YOUR website! HI, \n" ); document.write( "2x^4/3 + 5x^2/3 -13 =0 \n" ); document.write( ". \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:. \n" ); document.write( ". \n" ); document.write( " \n" ); document.write( ". \n" ); document.write( " \n" ); document.write( ". \n" ); document.write( " \n" ); document.write( ". \n" ); document.write( " \n" ); document.write( ". \n" ); document.write( " \n" ); document.write( ". \n" ); document.write( " \n" ); document.write( ". \n" ); document.write( " \n" ); document.write( "or\r \n" ); document.write( "\n" ); document.write( " |