+1124 Hi friends,
I may have been explained this before, but I just cannot get my head around this, no matter how I try, I just do not grasp this...may I ask just 10 min of your time to just explain this to me again, please..
The sum is: \(M= {{\sqrt{25-2x}} \over x}\)
Determine the value(s) of x for which M will be:
1.1) Real
1.2) Rasional
what confuses me is the WHY
WHY do they equate 25-2x greater and equal to zero for number 1?
for number 2, they find values for x, in this case 8 and 12. They plug them in and get answers...WHY 8 and 12?
ALSO, why not equate 25-2x = 0, or to anything else for number 2?
Rasional numbers are a part of REAL numbers, ....so why not treat them the same?
guys, I really would appreciate your assistance. Thank you all very much..
+128474 1) We cannot take the square root of a negative number and get a real number returned
Therfore.....the quantity under the radical MUST BE ≥ 0
So....
25 - 2x ≥ 0 add 2x to both sides
25 ≥ 2x divide both sides by 2
25/2 ≥ x which is the same as x ≤ 25/2
+128474 2) Any value of x that makes 25 - 2x ≥ 0 but NOT a perfect square will mean that a radical will still remain in the expression......and the remaining radical WILL NOT be a rational number....to see this
Let x = 2 and we have that
√[25 - (2)(2) ] / 2 = √21 / 2 and √21/2 is NOT RATIONAL
So......we only get a rational whenever x = 8 or x = 12 thusly
√[25 - 2(8) ] / 2 = √9 / 2 = 3/2 which IS rational
And
√ [25 - 2(12) ] / 2 = √1 /2 = 1/2 which is also rational
So...remember.....any positive integer under the radical that is not a perfect square will produce an irrational
Hope that helps !!!
+1124 CPill,
Thank you for your time, I am going to spend some time with examples of these types of questions, however I am sure the light will go on...
..I honestly do appreciate!!..have a blessed day...
