document.write( "Question 124289: Solve the following quadratic equation by completing the square. I get the first part then i get lost, please help me. x^2-6x-3=0 \n" ); document.write( "

Algebra.Com's Answer #91087 by jim_thompson5910(35256) $\"\"$ $\"About$

You can put this solution on YOUR website!

\n" ); document.write( " $\"y=x%5E2-6+x-3\"$ Start with the given equation\r
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\n" ); document.write( "\n" ); document.write( " $\"y%2B3=x%5E2-6+x\"$ Add $\"3\"$ to both sides\r
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\n" ); document.write( "\n" ); document.write( "Take half of the x coefficient $\"-6\"$ to get $\"-3\"$ (ie $\"%281%2F2%29%28-6%29=-3\"$ ).\r
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\n" ); document.write( "\n" ); document.write( "Now square $\"-3\"$ to get $\"9\"$ (ie $\"%28-3%29%5E2=%28-3%29%28-3%29=9\"$ )\r
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\n" ); document.write( "\n" ); document.write( " $\"y%2B3=x%5E2-6x%2B9-9\"$ Now add and subtract this value inside the parenthesis. Doing both the addition and subtraction of $\"9\"$ does not change the equation\r
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\n" ); document.write( "\n" ); document.write( " $\"y%2B3=%28x-3%29%5E2-9\"$ Now factor $\"x%5E2-6x%2B9\"$ to get $\"%28x-3%29%5E2\"$ \r
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\n" ); document.write( "\n" ); document.write( " $\"y=%28x-3%29%5E2-9-3\"$ Now add $\"%2B3\"$ to both sides to isolate y\r
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\n" ); document.write( "\n" ); document.write( " $\"y=%28x-3%29%5E2-12\"$ Combine like terms\r
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\n" ); document.write( "\n" ); document.write( " $\"%28x-3%29%5E2-12=0\"$ Now to solve for x, let $\"y=0\"$ \r
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\n" ); document.write( "\n" ); document.write( " $\"%28x-3%29%5E2=%2B12\"$ Add $\"12\"$ to both sides\r
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Take the square root of both sides\r
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Simplify $\"sqrt%2812%29\"$ to get $\"2sqrt%283%29\"$ \r
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Add $\"3\"$ to both sides\r
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\n" ); document.write( "\n" ); document.write( "So our answers are \r
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