document.write( "Question 201492This question is from textbook precalculus
\n" ); document.write( ": the square root of x minus 3 square root of x with an index of 4 minus 4 equals 0 \n" ); document.write( "
Algebra.Com's Answer #151702 by jsmallt9(3758)\"\" \"About 
You can put this solution on YOUR website!
The easiest way to do this is based on understanding that \"sqrt%28x%29+=+%28root%284%2C+x%29%29%5E2\". This is not hard to see if you understand fractional exponents: \"sqrt%28x%29+=+x%5E%281%2F2%29\" and \"root%284%2C+x%29+=+x%5E%281%2F4%29\". So \"%28root%284%2C+x%29%29%5E2+=+%28x%5E%281%2F4%29%29%5E2\". Using the appropriate rule for exposnents (\"%28a%5Eb%29%5Ec+=+a%5E%28b%2Ac%29\"

\n" ); document.write( "If we let \"y+=+root%284%2C+x%29\" then \"y%5E2+=+sqrt%28x%29\". Substituting these into
\n" ); document.write( "\"sqrt%28x%29+-3%2A%28root%284%2Cx%29%29+-4+=+0\" we get
\n" ); document.write( "\"y%5E2+-+3y+-+4+=+0\"
\n" ); document.write( "This equation can be easily solved by factoring:
\n" ); document.write( "\"%28y+-+4%29%2A%28y+%2B+1%29+=+0\"
\n" ); document.write( "Since a product of zero means one of the factors must be zero:
\n" ); document.write( "y-4=0 or y+1=0
\n" ); document.write( "Adding 4 to both sides of the first equation and subtracting 1 from both sides of the second we get:
\n" ); document.write( "y=4 or y=-1
\n" ); document.write( "Now we substitute back in for y:
\n" ); document.write( "\"root%284%2C+x%29+=+4\" or \"root%284%2C+x%29+=+-1\"
\n" ); document.write( "We reject the second equation because by definition \"root%284%2C+x%29\" must not be negative. To solve the first equation we raise both sides to the 4th power:
\n" ); document.write( "\"%28root%284%2C+x%29%29%5E4+=+4%5E4\" which gives:
\n" ); document.write( "x = 256\r
\n" ); document.write( "\n" ); document.write( "Checking:
\n" ); document.write( "\"sqrt%28256%29+-+3%2Aroot%284%2C+256%29+-+4+=+0\"
\n" ); document.write( "16 - 3*4 - 4 = 0
\n" ); document.write( "16 - 12 - 4 = 0
\n" ); document.write( "0 = 0 Check!!
\n" ); document.write( "
\n" );