document.write( "Question 936216: Please solve : log (k+2) + log (k-1) = 1 \n" ); document.write( "
Algebra.Com's Answer #569652 by srinivas.g(540)\"\" \"About 
You can put this solution on YOUR website!
log (k+2) + log (k-1) = 1 log 10 =1\r
\n" ); document.write( "\n" ); document.write( "log (k+2) + log (k-1) = log 10
\n" ); document.write( " formula : log a+log b = log(a*b)
\n" ); document.write( " log (k+2)*(k-1)=log 10
\n" ); document.write( " (k+2)(k-1)=10
\n" ); document.write( " k(k-1)+2(k-1)=10
\n" ); document.write( " \"+k%5E2%2Bk%2A%28-1%29%2B2%2Ak%2B2%2A%28-1%29++=10\"
\n" ); document.write( " \"+k%5E2-k%2B2k-2+=10\"
\n" ); document.write( " \"+k%5E2%2Bk-2=10\"
\n" ); document.write( " move 10 to the left
\n" ); document.write( " \"+k%5E2%2Bk-2-10=0\"
\n" ); document.write( " \"+k%5E2%2Bk-12=0\"
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Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation \"ak%5E2%2Bbk%2Bc=0\" (in our case \"1k%5E2%2B1k%2B-12+=+0\") has the following solutons:
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\n" ); document.write( " \"k%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca\"
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\n" ); document.write( " For these solutions to exist, the discriminant \"b%5E2-4ac\" should not be a negative number.
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\n" ); document.write( " First, we need to compute the discriminant \"b%5E2-4ac\": \"b%5E2-4ac=%281%29%5E2-4%2A1%2A-12=49\".
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\n" ); document.write( " Discriminant d=49 is greater than zero. That means that there are two solutions: \"+x%5B12%5D+=+%28-1%2B-sqrt%28+49+%29%29%2F2%5Ca\".
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\n" ); document.write( " \"k%5B1%5D+=+%28-%281%29%2Bsqrt%28+49+%29%29%2F2%5C1+=+3\"
\n" ); document.write( " \"k%5B2%5D+=+%28-%281%29-sqrt%28+49+%29%29%2F2%5C1+=+-4\"
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\n" ); document.write( " Quadratic expression \"1k%5E2%2B1k%2B-12\" can be factored:
\n" ); document.write( " \"1k%5E2%2B1k%2B-12+=+1%28k-3%29%2A%28k--4%29\"
\n" ); document.write( " Again, the answer is: 3, -4.\n" ); document.write( "Here's your graph:
\n" ); document.write( "\"graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B1%2Ax%2B-12+%29\"

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