document.write( "Question 38633: Let y = sqrt((x+1)(x+3)/(x+2).)Find all the real values of x for which y takes real
\n" ); document.write( "values. \n" ); document.write( "
Algebra.Com's Answer #24186 by AnlytcPhil(1806)\"\" \"About 
You can put this solution on YOUR website!
Let y = sqrt((x+1)(x+3)/(x+2).)Find all the real values of x for which\r\n" );
document.write( "y takes real values.\r\n" );
document.write( "\r\n" );
document.write( "      ____________\r\n" );
document.write( "_    / (x+1)(x+3)\r\n" );
document.write( " \  /  ——————————\r\n" );
document.write( "  \/      x+2\r\n" );
document.write( "\r\n" );
document.write( "A square root (or any even root) must have only positive numbers or zero\r\n" );
document.write( "under it.  So we must have \r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( " (x+1)(x+3)\r\n" );
document.write( " —————————— > 0\r\n" );
document.write( "    x+2\r\n" );
document.write( "\r\n" );
document.write( "The critical values are found by setting all the factors of the\r\n" );
document.write( "numerator and denominator = 0, and solving for x.  So the critical\r\n" );
document.write( "values are -1, -3, and -2\r\n" );
document.write( "\r\n" );
document.write( "Put these critical values on a number line:\r\n" );
document.write( "\r\n" );
document.write( "    -------o-----o-----o------------------\r\n" );
document.write( "    -4    -3    -2    -1     0     1     2    \r\n" );
document.write( "\r\n" );
document.write( "Pick a number less than the least critical value, say -4 and substitute\r\n" );
document.write( "it in\r\n" );
document.write( "\r\n" );
document.write( " (x+1)(x+3)\r\n" );
document.write( " ——————————\r\n" );
document.write( "    x+2\r\n" );
document.write( "\r\n" );
document.write( " (-4+1)(-4+3)   (-3)(-1)       3\r\n" );
document.write( " ———————————— = ————————— = - ——— \r\n" );
document.write( "    -4+2           -2          2\r\n" );
document.write( "\r\n" );
document.write( "This is negative, so we cannot shade left of -3\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "Next pick a number between -3 and -2, say -2.5 and substitute\r\n" );
document.write( "it in\r\n" );
document.write( "\r\n" );
document.write( " (x+1)(x+3)\r\n" );
document.write( " ——————————\r\n" );
document.write( "    x+2\r\n" );
document.write( "\r\n" );
document.write( " (-2.5+1)(-2.5+3)   (-1.5)(.5)    3\r\n" );
document.write( " ———————————————— = ————————— =  ——— \r\n" );
document.write( "      -2.5+2           -.5        2\r\n" );
document.write( "\r\n" );
document.write( "This is positive, so we MUST shade between -3 and 2, so now the\r\n" );
document.write( "number line becomes\r\n" );
document.write( "\r\n" );
document.write( "    -------o=====o-----o------------------\r\n" );
document.write( "    -4    -3    -2    -1     0     1     2\r\n" );
document.write( "\r\n" );
document.write( "Next pick a number between -2 and -1, say -1.5 and substitute\r\n" );
document.write( "it in\r\n" );
document.write( "\r\n" );
document.write( " (x+1)(x+3)\r\n" );
document.write( " ——————————\r\n" );
document.write( "    x+2\r\n" );
document.write( "\r\n" );
document.write( " (-1.5+1)(-1.5+3)   (-.5)(1.5)      3\r\n" );
document.write( " ———————————————— = ————————— =  - ——— \r\n" );
document.write( "      -1.5+2           .5           2\r\n" );
document.write( "\r\n" );
document.write( "This is nagative, so we do not shade between -2 and -1, so the\r\n" );
document.write( "number line is still \r\n" );
document.write( "\r\n" );
document.write( "    -------o=====o-----o------------------\r\n" );
document.write( "    -4    -3    -2    -1     0     1     2\r\n" );
document.write( "\r\n" );
document.write( "Next pick a number larger than -1, say 0 and substitute\r\n" );
document.write( "it in\r\n" );
document.write( "\r\n" );
document.write( " (x+1)(x+3)\r\n" );
document.write( " ——————————\r\n" );
document.write( "    x+2\r\n" );
document.write( "\r\n" );
document.write( " (0+1)(0+3)   (1)(5)        5\r\n" );
document.write( " —————————— = ————————— =  ——— \r\n" );
document.write( "    0+2          2          2\r\n" );
document.write( "\r\n" );
document.write( "This is positive, so we MUST shade to the right of -1, so the \r\n" );
document.write( "number line becomes\r\n" );
document.write( " \r\n" );
document.write( "\r\n" );
document.write( "    -------o=====o-----o==================>\r\n" );
document.write( "    -4    -3    -2    -1     0     1     2\r\n" );
document.write( "\r\n" );
document.write( "Finally we substitute the critical values themselves to see if they\r\n" );
document.write( "are solutions or not\r\n" );
document.write( "\r\n" );
document.write( "Substituting -3\r\n" );
document.write( "\r\n" );
document.write( " (x+1)(x+3)\r\n" );
document.write( " ——————————\r\n" );
document.write( "    x+2\r\n" );
document.write( "\r\n" );
document.write( " (-3+1)(-3+3)   (-2)(0)       \r\n" );
document.write( " ———————————— = ———————— = 0\r\n" );
document.write( "    -3+2           -1          \r\n" );
document.write( "\r\n" );
document.write( "Zero in a numerator is OK (as long as the denominator is not zero \r\n" );
document.write( "so -3 is a solution and so we place \"[\" where the \"o\" is to indicate\r\n" );
document.write( "that -3 is part of the solution set.\r\n" );
document.write( "\r\n" );
document.write( "    -------[=====o-----o==================>\r\n" );
document.write( "    -4    -3    -2    -1     0     1     2\r\n" );
document.write( "\r\n" );
document.write( "Substituting critical value -2\r\n" );
document.write( "\r\n" );
document.write( " (x+1)(x+3)\r\n" );
document.write( " ——————————\r\n" );
document.write( "    x+2\r\n" );
document.write( "\r\n" );
document.write( " (-2+1)(-2+3)   (-1)(1)       \r\n" );
document.write( " ———————————— = ———————— is not defined because of the 0 denominator\r\n" );
document.write( "    -2+2           0          \r\n" );
document.write( "\r\n" );
document.write( "so we place \")\" where the \"o\" is at -2 to indicate that it is not part\r\n" );
document.write( "of the solution set.\r\n" );
document.write( "\r\n" );
document.write( "    -------[=====)-----o==================>\r\n" );
document.write( "    -4    -3    -2    -1     0     1     2\r\n" );
document.write( "\r\n" );
document.write( "Substituting critical value -1\r\n" );
document.write( "\r\n" );
document.write( " (x+1)(x+3)\r\n" );
document.write( " ——————————\r\n" );
document.write( "    x+2\r\n" );
document.write( "\r\n" );
document.write( " (-1+1)(-1+3)   (0)(4)       \r\n" );
document.write( " ———————————— = —————— = 0\r\n" );
document.write( "    -1+2           1          \r\n" );
document.write( "\r\n" );
document.write( "Zero in a numerator is OK (as long as the denominator is not zero \r\n" );
document.write( "so -1 is a solution and so we place \"[\" where the \"o\" is to indicate\r\n" );
document.write( "that -1 is part of the solution set.\r\n" );
document.write( "\r\n" );
document.write( "    -------[=====)-----[==================>\r\n" );
document.write( "    -4    -3    -2    -1     0     1     2\r\n" );
document.write( "\r\n" );
document.write( "We imagine that there is an ¥ symbol on the \r\n" );
document.write( "far right:\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "    -------[=====)-----[==================> ¥\r\n" );
document.write( "    -4    -3    -2    -1     0     1     2\r\n" );
document.write( "\r\n" );
document.write( "So we construct our interval notation solution from the number line\r\n" );
document.write( "by writing the critical values inside the parentheses and brackets,\r\n" );
document.write( "and when we skip a section between, we write \"U\". So the answer is\r\n" );
document.write( "\r\n" );
document.write( "           [-3, -2) U [-1, ¥)\r\n" );
document.write( "\r\n" );
document.write( "We always put a \"(\" before -¥ and a \")\" after ¥  \r\n" );
document.write( "\r\n" );
document.write( "Edwin\r\n" );
document.write( "AnlytcPhil@aol.com
\n" ); document.write( "
\n" );