document.write( "Question 569282: solve the following radical equation: $\"sqrt+%28+3x%2B1+%29+-+sqrt+%28+2x-10+%29=+4+\"$ \n" ); document.write( "

Algebra.Com's Answer #367298 by AnlytcPhil(1806) $\"\"$ $\"About$

You can put this solution on YOUR website!

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document.write( "Isolate one radical term on one side:\r\n" );
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document.write( "                   = \r\n" );
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document.write( "                 = \r\n" );
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document.write( "     - 8 + 16 = 2x - 10\r\n" );
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document.write( "    3x + 1 - 8 + 16 = 2x - 10\r\n" );
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document.write( "        3x - 8 + 17 = 2x - 10\r\n" );
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document.write( "Isolate the radical term on one side:\r\n" );
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document.write( "                  x + 27 = 8\r\n" );
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document.write( "Square both sides:\r\n" );
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document.write( "               (x + 27)² = 8²\r\n" );
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document.write( "          x² + 54x + 729 = 64(3x + 1)\r\n" );
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document.write( "          x² + 54x + 729 = 192x + 64\r\n" );
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document.write( "Get 0 on the right:\r\n" );
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document.write( "         x² - 138x + 665 = 0         \r\n" );
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document.write( "Factor:\r\n" );
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document.write( "        (x - 5)(x - 133) = 0\r\n" );
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document.write( "Use the zero factor principle:\r\n" );
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document.write( "    x - 5 = 0        x - 133 = 0 \r\n" );
document.write( "        x = 5              x = 133\r\n" );
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document.write( "We much check all solution because squaring both sides\r\n" );
document.write( "often but not always introduces extraneous answers.\r\n" );
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document.write( "Checking x = 5 in the original:\r\n" );
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document.write( "4 - 0 = 4\r\n" );
document.write( "    4 = 4\r\n" );
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document.write( "So x = 5 is a solution.\r\n" );
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document.write( "Checking x = 133 in the original:\r\n" );
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document.write( "20 - 16 = 4\r\n" );
document.write( "      4 = 4\r\n" );
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document.write( "So x = 133 is also a solution.\r\n" );
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document.write( "Edwin

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