SOLUTION: Let y = sqrt((x+1)(x+3)/(x+2).)Find all the real values of x for which y takes real values.

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Question 38633: Let y = sqrt((x+1)(x+3)/(x+2).)Find all the real values of x for which y takes real
values.
Found 2 solutions by venugopalramana, AnlytcPhil:
Answer by venugopalramana(3286)   (Show Source): 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 y takes real
values.
FOR Y TO BE REAL (X+1)(X+3)/(X+2)SHOULD BE GREATER THAN OR EQUAL TO ZERO.
CASE 1......NR AND DR +VE
NR....(X+3)(X+1)SHOULD BE POSITIVE THAT IS
X>-1........I,OR......X<-3....................II
DR...(X+2) SHOULD BE +VE THAT IS
X>-2.............III
COMBINING I,II,III,WE GET FOR CASE 1.............X>-1........ANSWER SET 1
CASE 2....NR AND DR -VE.
NR.....(X+3)(X+1) SHOULD BE - VE.
THAT IS -3 DR....(X+2) SHOULD BE -VE....X<-2.............V
COMBINING IV AND V WE GET -3 SINCE NR COULD BE ZERO TOO,WE GET X=-3 AND -1 FOR THIS CASE.HENCE THE TOTAL SOLUTION SET IS
X=-3,X=-1,X>-1,-3 X>=-1....AND.....-3<=X<-2
Answer by AnlytcPhil(1806)   (Show Source): 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
y takes real values.

      ____________
_    / (x+1)(x+3)
 \  /  ——————————
  \/      x+2

A square root (or any even root) must have only positive numbers or zero
under it.  So we must have 


 (x+1)(x+3)
 —————————— > 0
    x+2

The critical values are found by setting all the factors of the
numerator and denominator = 0, and solving for x.  So the critical
values are -1, -3, and -2

Put these critical values on a number line:

    -------o-----o-----o------------------
    -4    -3    -2    -1     0     1     2    

Pick a number less than the least critical value, say -4 and substitute
it in

 (x+1)(x+3)
 ——————————
    x+2

 (-4+1)(-4+3)   (-3)(-1)       3
 ———————————— = ————————— = - ——— 
    -4+2           -2          2

This is negative, so we cannot shade left of -3


Next pick a number between -3 and -2, say -2.5 and substitute
it in

 (x+1)(x+3)
 ——————————
    x+2

 (-2.5+1)(-2.5+3)   (-1.5)(.5)    3
 ———————————————— = ————————— =  ——— 
      -2.5+2           -.5        2

This is positive, so we MUST shade between -3 and 2, so now the
number line becomes

    -------o=====o-----o------------------
    -4    -3    -2    -1     0     1     2

Next pick a number between -2 and -1, say -1.5 and substitute
it in

 (x+1)(x+3)
 ——————————
    x+2

 (-1.5+1)(-1.5+3)   (-.5)(1.5)      3
 ———————————————— = ————————— =  - ——— 
      -1.5+2           .5           2

This is nagative, so we do not shade between -2 and -1, so the
number line is still 

    -------o=====o-----o------------------
    -4    -3    -2    -1     0     1     2

Next pick a number larger than -1, say 0 and substitute
it in

 (x+1)(x+3)
 ——————————
    x+2

 (0+1)(0+3)   (1)(5)        5
 —————————— = ————————— =  ——— 
    0+2          2          2

This is positive, so we MUST shade to the right of -1, so the 
number line becomes
 

    -------o=====o-----o==================>
    -4    -3    -2    -1     0     1     2

Finally we substitute the critical values themselves to see if they
are solutions or not

Substituting -3

 (x+1)(x+3)
 ——————————
    x+2

 (-3+1)(-3+3)   (-2)(0)       
 ———————————— = ———————— = 0
    -3+2           -1          

Zero in a numerator is OK (as long as the denominator is not zero 
so -3 is a solution and so we place "[" where the "o" is to indicate
that -3 is part of the solution set.

    -------[=====o-----o==================>
    -4    -3    -2    -1     0     1     2

Substituting critical value -2

 (x+1)(x+3)
 ——————————
    x+2

 (-2+1)(-2+3)   (-1)(1)       
 ———————————— = ———————— is not defined because of the 0 denominator
    -2+2           0          

so we place ")" where the "o" is at -2 to indicate that it is not part
of the solution set.

    -------[=====)-----o==================>
    -4    -3    -2    -1     0     1     2

Substituting critical value -1

 (x+1)(x+3)
 ——————————
    x+2

 (-1+1)(-1+3)   (0)(4)       
 ———————————— = —————— = 0
    -1+2           1          

Zero in a numerator is OK (as long as the denominator is not zero 
so -1 is a solution and so we place "[" where the "o" is to indicate
that -1 is part of the solution set.

    -------[=====)-----[==================>
    -4    -3    -2    -1     0     1     2

We imagine that there is an ¥ symbol on the 
far right:


    -------[=====)-----[==================> ¥
    -4    -3    -2    -1     0     1     2

So we construct our interval notation solution from the number line
by writing the critical values inside the parentheses and brackets,
and when we skip a section between, we write "U". So the answer is

           [-3, -2) U [-1, ¥)

We always put a "(" before -¥ and a ")" after ¥  

Edwin
AnlytcPhil@aol.com
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