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Solving this inequality?

Solve this inequality for x:
(3+x)(6-x)<=0

Would I be right in saying that x=-3 or 6. Therefore, I would have said -3<=x<=6 but apparently that's wrong
Reply 1
Avatar for Jarred
Jarred
18
Original post by bobbricks
Solve this inequality for x:
(3+x)(6-x)<=0

Would I be right in saying that x=-3 or 6. Therefore, I would have said -3<=x<=6 but apparently that's wrong


What was your method in deciding this? or did you just go with this because it seemed right?

There are two methods you can go with, either produce a sketch or produce a sign diagram. I prefer the latter. I can explain these to you if don't know what they are.
Reply 2
Avatar for bobbricks
bobbricks
OP
19
Original post by Jarred
What was your method in deciding this? or did you just go with this because it seemed right?

There are two methods you can go with, either produce a sketch or produce a sign diagram. I prefer the latter. I can explain these to you if don't know what they are.


I drew a sketch and worked out that x had to be more than -3 and less than 6 for the y values to be less than 0...so I ended up with this: -3<=x<=6

Is that what you got?
(edited 10 years ago)
Reply 3
Avatar for james22
james22
16
Original post by bobbricks
I drew a sketch and worked out that x had to be more than -6 and less than 3 for the y values to be less than 0...so I ended up with this: -3<=x<=6

Is that what you got?


Was your sketch the wrong way up? Remember this quadratic has a negative x^2 term.
Reply 4
Avatar for bobbricks
bobbricks
OP
19
Original post by james22
Was your sketch the wrong way up? Remember this quadratic has a negative x^2 term.


Ah...that explains it (I did the sketch the positive way up :redface: ) Thanks :smile:
Reply 5
Avatar for Jarred
Jarred
18
Original post by bobbricks
I drew a sketch and worked out that x had to be more than -3 and less than 6 for the y values to be less than 0...so I ended up with this: -3<=x<=6

Is that what you got?


Correct method but the answer isn't quite there. As James suggested, your sketch could be the wrong way up. The function has a negative gradient if you expand it out, so it needs to be upside down to an ordinary quadratic.
Reply 6
Avatar for bobbricks
bobbricks
OP
19
Original post by james22
Was your sketch the wrong way up? Remember this quadratic has a negative x^2 term.



Original post by Jarred
Correct method but the answer isn't quite there. As James suggested, your sketch could be the wrong way up. The function has a negative gradient if you expand it out, so it needs to be upside down to an ordinary quadratic.


I've managed to get there thanks :biggrin:
(3+x)(6-x)<=0
This is slightly off topic but if I were to expand out the brackets I would get: -x2+3x-18<=0 which is equal to x2-3x+18>=0 ...However, that doesn't factorise- why?
Reply 7
Avatar for james22
james22
16
Original post by bobbricks
I've managed to get there thanks :biggrin:
(3+x)(6-x)<=0
This is slightly off topic but if I were to expand out the brackets I would get: -x2+3x-18<=0 which is equal to x2-3x+18>=0 ...However, that doesn't factorise- why?


You expanded it wrong, should be +18 not -18.
Reply 8
Avatar for shubhro
shubhro
1
Original post by bobbricks
Solve this inequality for x:
(3+x)(6-x)<=0

Would I be right in saying that x=-3 or 6. Therefore, I would have said -3<=x<=6 but apparently that's wrong


if (3 - x)(6 - x) <=0

this is only possible when
case 1. (3 - x) is -ve and (6 - x) is +ve, because -ve x +ve = -ve
case 2. (3 - x) is +ve and (6 - x) is -ve, because -ve x +ve = -ve

case 1.
3 - x <= 0 => x >= 3,
6 - x > = 0 => x =< 6

so, 3 =< x =< 6

similarly solve case 2.
(edited 10 years ago)
Reply 9
Avatar for shubhro
shubhro
1
if you are stuck, then you may wish to visit
http://ipm-mtse-olympiad-state-scholarship.blogspot.in/2013/10/range-of-expression.html#more
Reply 10
Avatar for bobbricks
bobbricks
OP
19
Ah, thanks guys, 10pm is certainly not the time to do maths! :tongue:

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