Another math question...Please tell me what I'm doing wrong...
[From: ] [author: ] [Date: 13-07-20] [Hit: ]
then it satisfies the equality.If you try xTherefore x>-6.If you try x>3, x=4, then it does not satisfyTherefore -6Its A.ANOTHER NOTE:When you substitute in -4 for x,......
next lets see lets test a # between -6 & - infinity, hows bout -8?
now if one plugs in - 8 they get (-2)(-11) = 22 so: 22 < 0 which is false
so that doesn't work.
-∞ ____X_______ -6 ___________ 0____3 _______
now lets try a between -6 & 3
welll hows bout 0?
which gives (6)(-3) = - 18
-18 < 0 so true
now if we want we can stop there since we've shown that it only works for values between -6 & 3 and our answers only give us that.
BUT for wholeness sake one could test a # greater than 3.
Anyways since it works for #s inclusive to -6 and 3 the answer is A.
You plugged in the wrong #s, that beign 6 & -3, remember to use the 0's (-6 and 3)
-
x^2 + 3x - 18 < 0
(x+6)(x-3) < 0
When you factored it you accounted the numbers as 6 and -3. X does NOT equal 6 and -3. In fact, it must cancel out the bracket so it must be -6 and 3.
If you try x>-6, x = -5, then it satisfies the equality.
If you try x<-6, x = -7, it does not satisfiy the equality.
Therefore x>-6.
If you try x>3, x=4, then it does not satisfy
Therefore -6
It's A.
ANOTHER NOTE:
When you substitute in -4 for x, it IS NOT -4^2. It's (-4)^2 making it +16 not -16.
-
After you factor the inequality as
(x+6)(x-3)
You put 6 and -3 on the number line; however, you should have put -6 and 3 on the number line. This is because those are your zeroes.
The zeroes are where the graph changes sign from + to -, or - to +.
Then you should pick points like -7, 0, and 4.
That should help.
-
U forgot an important fact. (x+6)(x-3) <0 if -6
keywords: tell,what,Please,Another,math,039,me,wrong,doing,question,Another math question...Please tell me what I'm doing wrong...
