document.write( "Question 84864: Solve by completing the square.
\n" ); document.write( " 4x^2 + 2x – 3 = 0
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Algebra.Com's Answer #61161 by jim_thompson5910(35256)\"\" \"About 
You can put this solution on YOUR website!
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Solved by pluggable solver: Completing the Square to Get a Quadratic into Vertex Form

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\n" ); document.write( " \"y=4+x%5E2%2B2+x-3\" Start with the given equation
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\n" ); document.write( " \"y%2B3=4+x%5E2%2B2+x\" Add \"3\" to both sides
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\n" ); document.write( " \"y%2B3=4%28x%5E2%2B%281%2F2%29x%29\" Factor out the leading coefficient \"4\"
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\n" ); document.write( " Take half of the x coefficient \"1%2F2\" to get \"1%2F4\" (ie \"%281%2F2%29%281%2F2%29=1%2F4\").
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\n" ); document.write( " Now square \"1%2F4\" to get \"1%2F16\" (ie \"%281%2F4%29%5E2=%281%2F4%29%281%2F4%29=1%2F16\")
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\n" ); document.write( " \"y%2B3=4%28x%5E2%2B%281%2F2%29x%2B1%2F16-1%2F16%29\" Now add and subtract this value inside the parenthesis. Doing both the addition and subtraction of \"1%2F16\" does not change the equation
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\n" ); document.write( " \"y%2B3=4%28%28x%2B1%2F4%29%5E2-1%2F16%29\" Now factor \"x%5E2%2B%281%2F2%29x%2B1%2F16\" to get \"%28x%2B1%2F4%29%5E2\"
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\n" ); document.write( " \"y%2B3=4%28x%2B1%2F4%29%5E2-4%281%2F16%29\" Distribute
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\n" ); document.write( " \"y%2B3=4%28x%2B1%2F4%29%5E2-1%2F4\" Multiply
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\n" ); document.write( " \"y=4%28x%2B1%2F4%29%5E2-1%2F4-3\" Now add \"%2B3\" to both sides to isolate y
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\n" ); document.write( " \"y=4%28x%2B1%2F4%29%5E2-13%2F4\" Combine like terms
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\n" ); document.write( " Now the quadratic is in vertex form \"y=a%28x-h%29%5E2%2Bk\" where \"a=4\", \"h=-1%2F4\", and \"k=-13%2F4\". Remember (h,k) is the vertex and \"a\" is the stretch/compression factor.
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\n" ); document.write( " Check:
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\n" ); document.write( " Notice if we graph the original equation \"y=4x%5E2%2B2x-3\" we get:
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\n" ); document.write( " \"graph%28500%2C500%2C-10%2C10%2C-10%2C10%2C4x%5E2%2B2x-3%29\" Graph of \"y=4x%5E2%2B2x-3\". Notice how the vertex is (\"-1%2F4\",\"-13%2F4\").
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\n" ); document.write( " Notice if we graph the final equation \"y=4%28x%2B1%2F4%29%5E2-13%2F4\" we get:
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\n" ); document.write( " \"graph%28500%2C500%2C-10%2C10%2C-10%2C10%2C4%28x%2B1%2F4%29%5E2-13%2F4%29\" Graph of \"y=4%28x%2B1%2F4%29%5E2-13%2F4\". Notice how the vertex is also (\"-1%2F4\",\"-13%2F4\").
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\n" ); document.write( " So if these two equations were graphed on the same coordinate plane, one would overlap another perfectly. So this visually verifies our answer.
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\n" ); document.write( "\n" ); document.write( "Since we have \r
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\n" ); document.write( "\n" ); document.write( "\"y=4%28x%2B1%2F4%29%5E2-13%2F4\"\r
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\n" ); document.write( "\n" ); document.write( "Set y=0 and solve for x\r
\n" ); document.write( "\n" ); document.write( "\"0=4%28x%2B1%2F4%29%5E2-13%2F4\"\r
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\n" ); document.write( "\n" ); document.write( "\"13%2F4=4%28x%2B1%2F4%29%5E2\" Add \"13%2F4\" to both sides\r
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\n" ); document.write( "\n" ); document.write( "\"%2813%2F4%29%281%2F4%29=%28x%2B1%2F4%29%5E2\" Multiply both sides by \"1%2F4\"\r
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\n" ); document.write( "\n" ); document.write( "\"0%2B-sqrt%2813%2F16%29=x%2B1%2F4\" Take the square root of both sides\r
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\n" ); document.write( "\n" ); document.write( "\"0%2B-sqrt%2813%29%2F4-1%2F4=x\" Subtract \"1%2F4\" from both sides\r
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\n" ); document.write( "\n" ); document.write( "So our answer is\r
\n" ); document.write( "\n" ); document.write( "\"x=%28-1%2B-sqrt%2813%29%29%2F4\"\r
\n" ); document.write( "\n" ); document.write( "which breaks down to \r
\n" ); document.write( "\n" ); document.write( "\"x=%28-1%2Bsqrt%2813%29%29%2F4\" or \"x=%28-1-sqrt%2813%29%29%2F4\"
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