Maths inequalities help please!

Original post by sienna2266

Hi guys,

So for this question I did:
-2x^2 +x +3≥ 0
2x^2 -x-3≤ 0
(2x-3)(x+1)≤ 0

Then I got stuck on drawing the graph.
This is what is shown in the mark scheme:

Attachment not found

However why can't the graph be:

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The roots remain the same because you just changed signs. However by changing signs you change the shape of the graph, you drew -f(x) rather than f(x). Hope that made sense. I tend to make that mistake too.

Reply 7

ddaanviiesl

8

Original post by Anonymouspsych

But you don't have to draw the negative quadratic graph. The OP has clearly taken all terms on the other side of the inequality and correctly drawn the corresponding graph. The problem can be solved either way...
It doesn't specify you have to draw a specific quadratic graph- "use a sketch of appropriate quadratic graphs..."

Fair enough, I would just recommend sketching the graph that is implied by the question as it is better practice for questions which involve having to get the sign correct or marks are lost. In this situation however this is not as important, obviously.

Reply 8

OP

Original post by RDKGames

Well it can't be like that because it's the reflected graph from the one asked by the question! They're not the same, they only share the same roots.
Not like it matters anyway, it's all just for the purposes of working out the same region.

Ahh thanks so much for replying :smile:

So I see what you mean there. Just to make sense of this. Here is the thread for a similar question that you helped me with ages ago: https://www.thestudentroom.co.uk/showthread.php?t=5154306&page=2
For this particular question(in the link above):
We or I :biggrin:

came to the conclusion that the graph can be sketched in 2 ways:
First way:

Attachment not found

However, now coming back to this, I think the second graph is wrong because I did x^2 +8<2x^2 +x +6 --> -x^2 -x +2<0 --> x^2 +x -2>0 --> (x-1)(x+2)>0
Clearly, the graph has a maximum point as it is -x^2 -x +2<0.

Could you please let me know your thoughts?

(edited 6 years ago)

OP

Hey guys, thanks for your replies!
Here is a very similar question which is rather confusing:

I've outlined my confusion in my reply to RDKGames.. Could someone please help with this one too?
So basically I understand they're asking you to "sketch" to just find out the solutions to the question but what I am specifically wondering about is the correct sketch of the graph. This one above is rather confusing
Please check my post above if possible- it would be much appreciated :smile:

(edited 6 years ago)

Reply 10

Original post by sienna2266

Hey guys, thanks for your replies!
Here is a very similar question which is rather confusing:

I've outlined my confusion in my reply to RDKGames.. Could someone please help with this one too?
So basically I understand they're asking you to "sketch" to just find out the solutions to the question but what I am specifically wondering about is the correct sketch of the graph. This one above is rather confusing
Please check my post above if possible- it would be much appreciated :smile:

If the coefficient of x in x^2 is positive it will form a u-shaped graph and if it's negative it will form an n-shaped graph.

Reply 11

OP

Original post by dont know it

If the coefficient of x in x^2 is positive it will form a u-shaped graph and if it's negative it will form an n-shaped graph.

Thanks but I am specifically talking about there being one graph on the axis not the one in the mark scheme. So this is not correct right:

It has to be the other way round? - so like n shaped?

Reply 12

Original post by sienna2266

Thanks but I am specifically talking about there being one graph on the axis not the one in the mark scheme. So this is not correct right:

It has to be the other way round? - so like n shaped?

No that is correct. By rearranging you get x^2+x-2>0. Since the coefficient of x is positive, you get a U-shape, so that's right.

Reply 13

h26

13

Original post by dont know it

No that is correct. By rearranging you get x^2+x-2>0. Since the coefficient of x is positive, you get a U-shape, so that's right.

" x^2 +8<2x^2 +x +6 --> -x^2 -x +2<0 --> x^2 +x -2>0 --> (x-1)(x+2)>0
Clearly, the graph has a maximum point as it is -x^2 -x +2<0. "

Just copy and pasted what OP did. I don't think the graph is correct cos this is the graph -x^2 -x +2<0 . The negative in front x^2 means it's got a maximum as opposed to a minimum point. The one OP has drawn is x^2 +x -2>0 and so the graph has been reflected which is wrong I think.
But that's just my gut instinct -hopefully some of the others will confirm :wink:

Reply 14

Original post by h26

" x^2 +8<2x^2 +x +6 --> -x^2 -x +2<0 --> x^2 +x -2>0 --> (x-1)(x+2)>0
Clearly, the graph has a maximum point as it is -x^2 -x +2<0. "

Just copy and pasted what OP did. I don't think the graph is correct cos this is the graph -x^2 -x +2<0 . The negative in front x^2 means it's got a maximum as opposed to a minimum point. The one OP has drawn is x^2 +x -2>0 and so the graph has been reflected which is wrong I think.
But that's just my gut instinct -hopefully some of the others will confirm :wink:

You can do it either way I think depending on however you rearranged it. Also, if you multiply that equation by -1, you get x^2+x-2>0.

Reply 15

OP

Original post by RDKGames

Well it can't be like that because it's the reflected graph from the one asked by the question! They're not the same, they only share the same roots.
Not like it matters anyway, it's all just for the purposes of working out the same region.

It would be extremely appreciated if you could please kindly reply back as you explain things so well.

Ahh thanks so much for replying :smile:

So I see what you mean there. Just to make sense of this. Here is the thread for a similar question that you helped me with ages ago: https://www.thestudentroom.co.uk/showthread.php?t=5154306&page=2
For this particular question(in the link above):
We or I :biggrin:

came to the conclusion that the graph can be sketched in 2 ways:
First way:

Second way:

Attachment not found

However, now coming back to this, I think the second graph is wrong because I did x^2 +8<2x^2 +x +6 --> -x^2 -x +2<0 --> x^2 +x -2>0 --> (x-1)(x+2)>0
Clearly, the graph has a maximum point as it is -x^2 -x +2<0.

Could you please let me know your thoughts?
I just felt I needed to refer back to this one as it contradicts the question in this thread

(edited 6 years ago)

Original post by sienna2266

It would be extremely appreciated if you could please kindly reply back as you explain things so well.

Ahh thanks so much for replying :smile:

So I see what you mean there. Just to make sense of this. Here is the thread for a similar question that you helped me with ages ago: https://www.thestudentroom.co.uk/showthread.php?t=5154306&page=2
For this particular question(in the link above):
We or I :biggrin:

came to the conclusion that the graph can be sketched in 2 ways:
First way:
However, now coming back to this, I think the second graph is wrong because I did x^2 +8<2x^2 +x +6 --> -x^2 -x +2<0 --> x^2 +x -2>0 --> (x-1)(x+2)>0
Clearly, the graph has a maximum point as it is -x^2 -x +2<0.

Could you please let me know your thoughts?
I just felt I needed to refer back to this one as it contradicts the question in this thread

It's not *wrong* it's just different graphs representing the same solutions to the inequality. If you want to sketch what the question lays out then yes you need to strictly follow the first way, but generally it doesn't matter.

Reply 17

OP

Original post by RDKGames

It's not *wrong* it's just different graphs representing the same solutions to the inequality. If you want to sketch what the question lays out then yes you need to strictly follow the first way, but generally it doesn't matter.

Thanks so much for replying. But if we ignore working out the solutions to the inequality and focus on drawing the correct sketch of the graph - doesn't that mean that the graph for way 2(alternative graph i thought of) should have a upside down U shaped graph? Btw I understand the graph shown in the mark scheme is correct. I am just wondering if the the alternative graph should be U shaped or upside down U shaped?
This is what I mean by upside down U shaped: and this is caused by -x^2 -x +2<0 -the solutions would be x<-2 or x>1

This is what I mean by U shaped graph: and this is caused by x^2 +x -2>0 - the solutions would be be x<-2 or x>1

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Both these graphs are from x^2 +8<2x^2 +x +6 and these graphs are not from the mark scheme -just an alternative graph i think could work
So I understand both these graphs yield the same solutions and they are both correct to use to work out the solutions. But what I am specifically asking is which one would be the correct representation/sketch of the graph x^2 +8<2x^2 +x +6?

Would be really really appreciated if you could please reply back. Thank you so much!

(edited 6 years ago)

Original post by sienna2266

But what I am specifically asking is which one would be the correct representation/sketch of the graph x^2 +8<2x^2 +x +6?

Would be really really appreciated if you could please reply back. Thank you so much!

Neither or both, whichever way you want to think about it. But the point is that one does not take precedence over the other, they are both equally as fine in order to answer the original inequality and both are fine to sketch in order to answer it.

Reply 19

OP