document.write( "Question 107967: solve the equation by completing the square
\n" ); document.write( "\"x%5E2%2B4x-1=0\"
\n" ); document.write( "\"x%5E2%2B6x-4=0\"
\n" ); document.write( "\"x%5E2-2x-5=0\"
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Algebra.Com's Answer #78679 by MathLover1(20849)\"\" \"About 
You can put this solution on YOUR website!

\n" ); document.write( "\"x%5E2%2B4x-1=0\"\r
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Solved by pluggable solver: COMPLETING THE SQUARE solver for quadratics
Read this lesson on completing the square by prince_abubu, if you do not know how to complete the square.
Let's convert \"1x%5E2%2B4x%2B-1=0\" to standard form by dividing both sides by 1:
\n" ); document.write( "We have: \"1x%5E2%2B4x%2B-1=0\". \n" ); document.write( "What we want to do now is to change this equation to a complete square \"%28x%2Bsomenumber%29%5E2+%2B+othernumber\". How can we find out values of somenumber and othernumber that would make it work?
\n" ); document.write( "Look at \"%28x%2Bsomenumber%29%5E2\": \"%28x%2Bsomenumber%29%5E2+=+x%5E2%2B2%2Asomenumber%2Ax+%2B+somenumber%5E2\". Since the coefficient in our equation \"1x%5E2%2Bhighlight_red%28+4%29+%2A+x%2B-1=0\" that goes in front of x is 4, we know that 4=2*somenumber, or \"somenumber+=+4%2F2\". So, we know that our equation can be rewritten as \"%28x%2B4%2F2%29%5E2+%2B+othernumber\", and we do not yet know the other number.
\n" ); document.write( "We are almost there. Finding the other number is simply a matter of not making too many mistakes. We need to find 'other number' such that \"%28x%2B4%2F2%29%5E2+%2B+othernumber\" is equivalent to our original equation \"1x%5E2%2B4x%2Bhighlight_green%28+-1+%29=0\".
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\n" ); document.write( " The highlighted red part must be equal to -1 (highlighted green part).
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\n" ); document.write( " \"4%5E2%2F4+%2B+othernumber+=+-1\", or \"othernumber+=+-1-4%5E2%2F4+=+-5\".
\n" ); document.write( "So, the equation converts to \"%28x%2B4%2F2%29%5E2+%2B+-5+=+0\", or \"%28x%2B4%2F2%29%5E2+=+5\".
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\n" ); document.write( " Our equation converted to a square \"%28x%2B4%2F2%29%5E2\", equated to a number (5).
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\n" ); document.write( " Since the right part 5 is greater than zero, there are two solutions:
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\n" ); document.write( " \"system%28+%28x%2B4%2F2%29+=+%2Bsqrt%28+5+%29%2C+%28x%2B4%2F2%29+=+-sqrt%28+5+%29+%29\"
\n" ); document.write( " , or
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\n" ); document.write( " \"system%28+%28x%2B4%2F2%29+=+2.23606797749979%2C+%28x%2B4%2F2%29+=+-2.23606797749979+%29\"
\n" ); document.write( " \"system%28+x%2B4%2F2+=+2.23606797749979%2C+x%2B4%2F2+=+-2.23606797749979+%29\"
\n" ); document.write( " \"system%28+x+=+2.23606797749979-4%2F2%2C+x+=+-2.23606797749979-4%2F2+%29\"
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\n" ); document.write( " \"system%28+x+=+0.23606797749979%2C+x+=+-4.23606797749979+%29\"
\n" ); document.write( "Answer: x=0.23606797749979, -4.23606797749979.\n" ); document.write( "

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\n" ); document.write( "\n" ); document.write( "\"x%5E2%2B6x-4=0\"\r
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Solved by pluggable solver: COMPLETING THE SQUARE solver for quadratics
Read this lesson on completing the square by prince_abubu, if you do not know how to complete the square.
Let's convert \"1x%5E2%2B6x%2B-4=0\" to standard form by dividing both sides by 1:
\n" ); document.write( "We have: \"1x%5E2%2B6x%2B-4=0\". \n" ); document.write( "What we want to do now is to change this equation to a complete square \"%28x%2Bsomenumber%29%5E2+%2B+othernumber\". How can we find out values of somenumber and othernumber that would make it work?
\n" ); document.write( "Look at \"%28x%2Bsomenumber%29%5E2\": \"%28x%2Bsomenumber%29%5E2+=+x%5E2%2B2%2Asomenumber%2Ax+%2B+somenumber%5E2\". Since the coefficient in our equation \"1x%5E2%2Bhighlight_red%28+6%29+%2A+x%2B-4=0\" that goes in front of x is 6, we know that 6=2*somenumber, or \"somenumber+=+6%2F2\". So, we know that our equation can be rewritten as \"%28x%2B6%2F2%29%5E2+%2B+othernumber\", and we do not yet know the other number.
\n" ); document.write( "We are almost there. Finding the other number is simply a matter of not making too many mistakes. We need to find 'other number' such that \"%28x%2B6%2F2%29%5E2+%2B+othernumber\" is equivalent to our original equation \"1x%5E2%2B6x%2Bhighlight_green%28+-4+%29=0\".
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\n" ); document.write( " The highlighted red part must be equal to -4 (highlighted green part).
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\n" ); document.write( " \"6%5E2%2F4+%2B+othernumber+=+-4\", or \"othernumber+=+-4-6%5E2%2F4+=+-13\".
\n" ); document.write( "So, the equation converts to \"%28x%2B6%2F2%29%5E2+%2B+-13+=+0\", or \"%28x%2B6%2F2%29%5E2+=+13\".
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\n" ); document.write( " Our equation converted to a square \"%28x%2B6%2F2%29%5E2\", equated to a number (13).
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\n" ); document.write( " Since the right part 13 is greater than zero, there are two solutions:
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\n" ); document.write( " \"system%28+%28x%2B6%2F2%29+=+%2Bsqrt%28+13+%29%2C+%28x%2B6%2F2%29+=+-sqrt%28+13+%29+%29\"
\n" ); document.write( " , or
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\n" ); document.write( " \"system%28+%28x%2B6%2F2%29+=+3.60555127546399%2C+%28x%2B6%2F2%29+=+-3.60555127546399+%29\"
\n" ); document.write( " \"system%28+x%2B6%2F2+=+3.60555127546399%2C+x%2B6%2F2+=+-3.60555127546399+%29\"
\n" ); document.write( " \"system%28+x+=+3.60555127546399-6%2F2%2C+x+=+-3.60555127546399-6%2F2+%29\"
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\n" ); document.write( " \"system%28+x+=+0.605551275463989%2C+x+=+-6.60555127546399+%29\"
\n" ); document.write( "Answer: x=0.605551275463989, -6.60555127546399.\n" ); document.write( "

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\n" ); document.write( "\n" ); document.write( "\"x%5E2-2x-5=0\"\r
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Solved by pluggable solver: COMPLETING THE SQUARE solver for quadratics
Read this lesson on completing the square by prince_abubu, if you do not know how to complete the square.
Let's convert \"1x%5E2%2B-2x%2B-5=0\" to standard form by dividing both sides by 1:
\n" ); document.write( "We have: \"1x%5E2%2B-2x%2B-5=0\". \n" ); document.write( "What we want to do now is to change this equation to a complete square \"%28x%2Bsomenumber%29%5E2+%2B+othernumber\". How can we find out values of somenumber and othernumber that would make it work?
\n" ); document.write( "Look at \"%28x%2Bsomenumber%29%5E2\": \"%28x%2Bsomenumber%29%5E2+=+x%5E2%2B2%2Asomenumber%2Ax+%2B+somenumber%5E2\". Since the coefficient in our equation \"1x%5E2%2Bhighlight_red%28+-2%29+%2A+x%2B-5=0\" that goes in front of x is -2, we know that -2=2*somenumber, or \"somenumber+=+-2%2F2\". So, we know that our equation can be rewritten as \"%28x%2B-2%2F2%29%5E2+%2B+othernumber\", and we do not yet know the other number.
\n" ); document.write( "We are almost there. Finding the other number is simply a matter of not making too many mistakes. We need to find 'other number' such that \"%28x%2B-2%2F2%29%5E2+%2B+othernumber\" is equivalent to our original equation \"1x%5E2%2B-2x%2Bhighlight_green%28+-5+%29=0\".
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\n" ); document.write( " The highlighted red part must be equal to -5 (highlighted green part).
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\n" ); document.write( " \"-2%5E2%2F4+%2B+othernumber+=+-5\", or \"othernumber+=+-5--2%5E2%2F4+=+-6\".
\n" ); document.write( "So, the equation converts to \"%28x%2B-2%2F2%29%5E2+%2B+-6+=+0\", or \"%28x%2B-2%2F2%29%5E2+=+6\".
\n" ); document.write( "
\n" ); document.write( " Our equation converted to a square \"%28x%2B-2%2F2%29%5E2\", equated to a number (6).
\n" ); document.write( "
\n" ); document.write( " Since the right part 6 is greater than zero, there are two solutions:
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\n" ); document.write( " \"system%28+%28x%2B-2%2F2%29+=+%2Bsqrt%28+6+%29%2C+%28x%2B-2%2F2%29+=+-sqrt%28+6+%29+%29\"
\n" ); document.write( " , or
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\n" ); document.write( " \"system%28+%28x%2B-2%2F2%29+=+2.44948974278318%2C+%28x%2B-2%2F2%29+=+-2.44948974278318+%29\"
\n" ); document.write( " \"system%28+x%2B-2%2F2+=+2.44948974278318%2C+x%2B-2%2F2+=+-2.44948974278318+%29\"
\n" ); document.write( " \"system%28+x+=+2.44948974278318--2%2F2%2C+x+=+-2.44948974278318--2%2F2+%29\"
\n" ); document.write( "
\n" ); document.write( " \"system%28+x+=+3.44948974278318%2C+x+=+-1.44948974278318+%29\"
\n" ); document.write( "Answer: x=3.44948974278318, -1.44948974278318.\n" ); document.write( "

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