document.write( "Question 1131704: Solve 6x⁴ - 35x³ + 62x² - 35x + 6 = 0 \n" ); document.write( "
Algebra.Com's Answer #748388 by ikleyn(52851)\"\" \"About 
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\n" ); document.write( "\n" ); document.write( "            There is a  SPECIAL  PROCEDURE  in algebra to solve such equations.  It is presented below.\r
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document.write( "\"6x%5E4+-+35x%5E3+%2B+62x%5E2+-+35x+%2B+6\" = 0      (1)\r\n" );
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document.write( "It follows from the equation that x= 0 IS NOT the root.\r\n" );
document.write( "So, we can divide both sides by  \"x%5E2\"  without loosing the roots. In this way, you will get an equivalent equation\r\n" );
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document.write( "\"6x%5E2+-+35x+%2B+62+-+35%2A%281%2Fx%29+%2B+6%2A%281%2Fx%5E2%29\" = 0.\r\n" );
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document.write( "Group and re-write it equivalently in the form\r\n" );
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document.write( "\"6%2A%28x%5E2+%2B+2+%2B+1%2Fx%5E2%29\" - \"%2835x+%2B+35%2A%281%2Fx%29%29\" + 50 = 0,    or\r\n" );
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document.write( "\"6%2A%28x%2B1%2Fx%29%5E2\" - \"35%2A%28x%2B1%2Fx%29\" + 50 = 0.     (2)\r\n" );
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document.write( "Introduce new variable  u = \"x\" + \"1%2Fx\".  Then equation (2) takes a form\r\n" );
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document.write( "\"6u%5E2+-+35u+%2B+50\" = 0.\r\n" );
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document.write( "Solve this quadratic equation using the quadratic formula\r\n" );
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document.write( "\"u%5B1%2C2%5D\" = \"35+%2B-+sqrt%2835%5E2+-+4%2A6%2A50%29%29%2F%282%2A6%29\" = \"%2835+%2B-+sqrt%2825%29%29%2F12\".\r\n" );
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document.write( "The two roots are  \r\n" );
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document.write( "    \"u%5B1%5D\" = \"%2835+%2B+sqrt%2825%29%29%2F12\" = \"%2835+%2B+5%29%2F12\" = \"40%2F12\" = \"10%2F3\"  and\r\n" );
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document.write( "    \"u%5B2%5D\" = \"%2835+-+sqrt%2825%29%29%2F12\" = \"%2835+-+5%29%2F12\" = \"30%2F12\" = \"10%2F4\".\r\n" );
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document.write( "Now, to find x,  we need to solve two equations\r\n" );
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document.write( "    a)  \"x\" + \"1%2Fx\" = \"10%2F3\"   and  b)  \"x\" + \"1%2Fx\" = \"10%2F4\".\r\n" );
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document.write( "Case a).  \"x\" + \"1%2Fx\" = \"10%2F3\"\r\n" );
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document.write( "          \"3x%5E2+-+10x+%2B+3\" = 0\r\n" );
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document.write( "          \"x%5B1%2C2%5D\" = \"%2810+%2B-+sqrt%28%28-10%29%5E2+-+4%2A3%2A3%29%29%2F%282%2A3%29\" = \"%2810+%2B-+sqrt%2864%29%29%2F6\" = \"%2810+%2B-+8%29%2F6\".\r\n" );
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document.write( "          So, the two roots are  \"x%5B1%5D\" = \"%2810+%2B+8%29%2F6\" = 3  and  \"x%5B2%5D\" = \"%2810+-+8%29%2F6\" = \"1%2F3\".\r\n" );
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document.write( "Case b).  \"x\" + \"1%2Fx\" = \"10%2F4\"\r\n" );
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document.write( "          \"4x%5E2+-+10x+%2B+4\" = 0\r\n" );
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document.write( "          \"x%5B3%2C4%5D\" = \"%2810+%2B-+sqrt%28%28-10%29%5E2+-+4%2A4%2A4%29%29%2F%282%2A4%29\" = \"%2810+%2B-+sqrt%2836%29%29%2F6\" = \"%2810+%2B-+6%29%2F8\".\r\n" );
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document.write( "          So, the two roots are  \"x%5B3%5D\" = \"%2810+%2B+6%29%2F8\" = 2  and  \"x%5B4%5D\" = \"%2810+-+6%29%2F8\" = \"1%2F2\".\r\n" );
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document.write( "ANSWER.  The four roots are   \"1%2F3\",  \"1%2F2\",  2  and  3.\r\n" );
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document.write( "    From this post learn on how to solve palindromic equations.\r\n" );
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document.write( "    Every palindromic equation of the degree 4 can be solved in this way.\r\n" );
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document.write( "    a)  Divide both sides by  \"x%5E2\";\r\n" );
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document.write( "    b)  Introduce new variable  u = x + \"1%2Fx\";\r\n" );
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document.write( "    c)  Reduce the equation to a quadratic equation relative new variable u  and solve it getting two roots  \"u%5B1%5D\"  and  \"u%5B2%5D\";\r\n" );
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document.write( "    d)  Then find  x  by solving two equations  \"x%2B1%2Fx\" = \"u%5B1%5D\"  and  \"x%2B1%2Fx\" = \"u%5B2%5D\".\r\n" );
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document.write( "    Every palindromic equation of the degree 4 can be solved in this way.\r\n" );
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