document.write( "Question 182730: I need help please, I could not find an example in the book.\r
\n" ); document.write( "\n" ); document.write( "Transform the function f(x)=x^2-10x+32 to the form f(x)=c(x-h)^2+k, where c,h,and k are constants, by completing the square.\r
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Algebra.Com's Answer #137197 by vleith(2983)\"\" \"About 
You can put this solution on YOUR website!
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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-10x%2B32=0\" to standard form by dividing both sides by 1:
\n" ); document.write( "We have: \"1x%5E2%2B-10x%2B32=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+-10%29+%2A+x%2B32=0\" that goes in front of x is -10, we know that -10=2*somenumber, or \"somenumber+=+-10%2F2\". So, we know that our equation can be rewritten as \"%28x%2B-10%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-10%2F2%29%5E2+%2B+othernumber\" is equivalent to our original equation \"1x%5E2%2B-10x%2Bhighlight_green%28+32+%29=0\".
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\n" ); document.write( " The highlighted red part must be equal to 32 (highlighted green part).
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\n" ); document.write( " \"-10%5E2%2F4+%2B+othernumber+=+32\", or \"othernumber+=+32--10%5E2%2F4+=+7\".
\n" ); document.write( "So, the equation converts to \"%28x%2B-10%2F2%29%5E2+%2B+7+=+0\", or \"%28x%2B-10%2F2%29%5E2+=+-7\".
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\n" ); document.write( " Our equation converted to a square \"%28x%2B-10%2F2%29%5E2\", equated to a number (-7).
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\n" ); document.write( " There is no number whose square can be negative. So, there is no solution to this equation
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