document.write( "Question 36935: Please solve and show all work.\r
\n" ); document.write( "\n" ); document.write( "I hope I enter the radical sign in correctly ( I am new to the formula plotting system), but if I do not the radical goes over x + 12. Please let me know if I entered correctly!\r
\n" ); document.write( "\n" ); document.write( "x= \"sqrt%28x+%2B+12%29\"
\n" ); document.write( "

Algebra.Com's Answer #22692 by venugopalramana(3286)\"\" \"About 
You can put this solution on YOUR website!
Please solve and show all work.
\n" ); document.write( "I hope I enter the radical sign in correctly ( I am new to the formula plotting system), but if I do not the radical goes over x + 12. Please let me know if I entered correctly!
\n" ); document.write( "x= sqrt(x + 12)
\n" ); document.write( "YOU DID VERY WELL.DONT WORRY EVEN IF YOU DONT KNOW TYPING THESE THINGS.WRITE THE PROBLEM AS YOU READ WITH BRACKETS ETC...THAT WILL DO ...HERE NOW...
\n" ); document.write( "SQUARE BOTH SIDES
\n" ); document.write( "X^2={SQRT(X+12)}^2=X+12
\n" ); document.write( "X^2-X-12=0
\n" ); document.write( "X^2-4X+3X-12=0
\n" ); document.write( "X(X-4)+3(X-4)=0
\n" ); document.write( "(X-4)(X+3)=0
\n" ); document.write( "X-4=0...THAT IS X=4
\n" ); document.write( "CHECK
\n" ); document.write( "4=SQRT(4+12)=SQRT(16)=4...OK
\n" ); document.write( "X+3=0...THAT IS X=-3...
\n" ); document.write( "CHECK
\n" ); document.write( "-3=SQRT(-3+12)=SQRT(9)=3..IN MATHS WE TAKE THE POSITIVE VALUE OF SQUARE ROOT IN GENERAL..SO WE DO NOT ACCEPT THIS SOLUTION,THOUGH SQRTOF 9 COULD BE -3 ALSO.
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