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B it is
(1) is insuff
Because x^2-7x>=0 means that x>=7 or x<=0
x(x-5)(x+2)=0 so x could be 0;5;-2
(1) only we get x= 0 (not negative) or -2 (negative)
So B is correct (x can get only one value 5 )
Fig can you elaborate on the logic of your answer ( Z level students teaching style plz?
thanks in advance
At least, I can try
Since x(x-5)(x+2)=0, then x could be -2, 0 or 5.
If we have a close look on these possible solutions, we can observe that 1 is negative, 1 is without sign and 1 is positive.
To answer the problem quesion :'is x negative?', we could have these groups of solutions:
> groupe 1 : x = 5 (Not neg)
> groupe 2 : x = 5 or x = 0 (Not neg)
> groupe 3 : x = 0 (Not neg)
> groupe 4 : x = -2 (Neg)
Now, we could look at the statments. These statments have normally to give us 1 of the above group in order to respond all answers except (E).
Stat1: x^2 - 7x >= 0
Thus, x*(x-7) >=0
<=> x <= 0 or x >= 7
The bold inequality shows us that x is negative or 0. It's not match to 1 of the 4 groups definied to answer the question. -2 is negative but 0 is neither negative nor positive.
INSUFF
Stat2: x^2 - 2x - 15>=0
Thus, (x-5)(x+3) >=0
<=> x >= 5 or x <= -3
Bingo, -2 and 0 are out the possible values of x while x=5 (groupe 1) is contained in the bold inequality.
SUFF
Getting bk to the inequality basics,...i'm awfully weak in inequalities...
thus, (x-5)(x+3) >=0 <=> x >= 5 or x <= -3
does x+3 >=0, not mean tht x>= -3( when u subtract 3 on both sides...)
please clear this doubt for me.....it wud help me a lot..coz inequality probs always pull me down.. _________________
ARISE AWAKE AND REST NOT UNTIL THE GOAL IS ACHIEVED
Getting bk to the inequality basics,...i'm awfully weak in inequalities... thus, (x-5)(x+3) >=0 <=> x >= 5 or x <= -3
does x+3 >=0, not mean tht x>= -3( when u subtract 3 on both sides...) please clear this doubt for me.....it wud help me a lot..coz inequality probs always pull me down..
Actually, the equation (x-5)(x+3) = x^2 - 2x - 15 = a*x^2 + b*x + c has the sign of :
o a (1 here) when x is not between the 2 roots, thus > 0
o -a (-1) when x is between the 2 roots, thus < 0
By the way, we have so:
(x-5)(x+3) >=0
<=> x >= 5 or x <= -3
Getting bk to the inequality basics,...i'm awfully weak in inequalities... thus, (x-5)(x+3) >=0 <=> x >= 5 or x <= -3
does x+3 >=0, not mean tht x>= -3( when u subtract 3 on both sides...) please clear this doubt for me.....it wud help me a lot..coz inequality probs always pull me down..
Actually, the equation (x-5)(x+3) = x^2 - 2x - 15 = a*x^2 + b*x + c has the sign of : o a (1 here) when x is not between the 2 roots, thus > 0 o -a (-1) when x is between the 2 roots, thus < 0
By the way, we have so: (x-5)(x+3) >=0 <=> x >= 5 or x <= -3
Fig - Thanks for the explanation. I kinda get it but am still a bit confused. How do you get x<=-3 ...when I subtarct three from x+3>=0 I get x>=-3. Thanks very much in advance.
Getting bk to the inequality basics,...i'm awfully weak in inequalities... thus, (x-5)(x+3) >=0 <=> x >= 5 or x <= -3
does x+3 >=0, not mean tht x>= -3( when u subtract 3 on both sides...) please clear this doubt for me.....it wud help me a lot..coz inequality probs always pull me down..
Actually, the equation (x-5)(x+3) = x^2 - 2x - 15 = a*x^2 + b*x + c has the sign of : o a (1 here) when x is not between the 2 roots, thus > 0 o -a (-1) when x is between the 2 roots, thus < 0
By the way, we have so: (x-5)(x+3) >=0 <=> x >= 5 or x <= -3
Fig - Thanks for the explanation. I kinda get it but am still a bit confused. How do you get x<=-3 ...when I subtarct three from x+3>=0 I get x>=-3. Thanks very much in advance.
Well, I can try
To seach (x-5)(x+3) >=0 is similar to study the sign of the function f(x) = (x-5)(x+3).
A great tool that Maths gives us is the table of signs. Once a fonction is factorized in "famous" forms of basic functions, such as a*x+b, we decompose the function and use the properties of the multiplication to determine what values of x give a positive sign and what values of x give a negative sign to the studied function f(x).
This table of signs is fast to draw and brings to the solution
I attached u the resulting table. To find the result signs for f(x), we multiply vertically + and - as they are +1 and -1.
So, when x <= -3 : sign(f(x)) = sign( (x-5)(x+3) ) = (-1)*(-1) = +1
I also join u the draw of the function f(x). U will visually remark that f(x) is positive on values of x not between the 2 roots.
This result is linked to my former post. The sign of a*x^2 + b*x + c
Attachments
Graph.jpg [ 31.16 KiB | Viewed 663 times ]
Signs-Table.jpg [ 14.17 KiB | Viewed 660 times ]
Last edited by Fig on 11 Oct 2006, 23:49, edited 2 times in total.
Thanks so much FIG for the great explanation. I need to take some time to let it sink in and try some more problems. Great graphs..I really appreciate it.