document.write( "Question 345185: solve with rational exponents then check solution...???...\r
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\n" ); document.write( "\n" ); document.write( "(x^2-x-4)^3/4-2=6 \n" ); document.write( "
Algebra.Com's Answer #246882 by jsmallt9(3758)\"\" \"About 
You can put this solution on YOUR website!
\"%28x%5E2-x-4%29%5E%283%2F4%29-2=6\"
\n" ); document.write( "First we'll isolate the exponentiated expression by adding 2 to each side:
\n" ); document.write( "\"%28x%5E2-x-4%29%5E%283%2F4%29=8\"
\n" ); document.write( "Now we want to get rid of that exponent. Actually evyerthing has an exponent. But we don't see the exponent if it is a 1. So when we say \"get rid of an exponent\", we really mean \"turn the exponent into a 1\". So how can we turn the 3/4 into a one? Any number (except 0) can be turned into a 1 by multiplying it by its reciprocal. So if we can figure out an operation that will allow us to multiply the exponent, 3/4, by its reciprocal, 4/3 then the exponent will \"disappear\". Fortunately we have a rule that says if you raise a power to a power then multiply the exponents. So all we need to do is raise both sides of the equation to the 4/3 power!
\n" ); document.write( "\"%28%28x%5E2-x-4%29%5E%283%2F4%29%29%5E%284%2F3%29=%288%29%5E%284%2F3%29\"
\n" ); document.write( "On the left the exponent turns into a 1 and disappears. On the right we need to simplify \"8%5E%284%2F3%29\". From what we should know about fractional exponents: \"8%5E%284%2F3%29+=+root%283%2C+8%5E4%29+=+%28root%283%2C+8%29%29%5E4\". Since 8 is a perfect cube I will use the second expression. (Don't worry, the first expression still works. But it is harder to simplify.) So now our equation is:
\n" ); document.write( "\"x%5E2-x-4+=+%28root%283%2C+8%29%29%5E4\"
\n" ); document.write( "\"x%5E2-x-4+=+%282%29%5E4\"
\n" ); document.write( "\"x%5E2-x-4+=+16\"
\n" ); document.write( "At this point we have a quadratic equation to solve. So we want one side to be zero. Subtracting 16 from each side will work:
\n" ); document.write( "\"x%5E2-x-20+=+0\"
\n" ); document.write( "Now we factor (or use the Quadratic Formula). This factors fairly easily:
\n" ); document.write( "\"%28x%2B4%29%28x-5%29+=+0\"
\n" ); document.write( "By the Zero Product property we know that this (or any) product can be zero only if one (or more) of the factors is zero. So:
\n" ); document.write( "\"x%2B4+=+0\" or \"x-5+=+0\"
\n" ); document.write( "Solving these we get:
\n" ); document.write( "\"x+=+-4\" or \"x+=+5\"

\n" ); document.write( "Now we'll check our answers, using the original equation:
\n" ); document.write( "\"%28x%5E2-x-4%29%5E%283%2F4%29-2=6\"
\n" ); document.write( "Checking x = -4:
\n" ); document.write( "\"%28%28-4%29%5E2-%28-4%29-4%29%5E%283%2F4%29-2=6\"
\n" ); document.write( "\"%2816-%28-4%29-4%29%5E%283%2F4%29-2=6\"
\n" ); document.write( "\"%2816%2B4-4%29%5E%283%2F4%29-2=6\"
\n" ); document.write( "\"%2816%29%5E%283%2F4%29-2=6\"
\n" ); document.write( "\"%28root%284%2C+16%29%29%5E3+-+2+=+6\"
\n" ); document.write( "\"%282%29%5E3+-+2+=+6\"
\n" ); document.write( "\"8+-+2+=+6\" Check!
\n" ); document.write( "Checking x = 5:
\n" ); document.write( "\"%28%285%29%5E2-%285%29-4%29%5E%283%2F4%29-2=6\"
\n" ); document.write( "\"%2825-%285%29-4%29%5E%283%2F4%29-2=6\"
\n" ); document.write( "\"%2816%29%5E%283%2F4%29-2=6\"
\n" ); document.write( "\"%28root%284%2C+16%29%29%5E3+-+2+=+6\"
\n" ); document.write( "\"%282%29%5E3+-+2+=+6\"
\n" ); document.write( "\"8+-+2+=+6\" Check!

\n" ); document.write( "So there are two solutions to the original equation: x = -4 and x = 5. \n" ); document.write( "
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