The Student Room Group

Inequalities and descriminant- c1

Slightly confused as to when I do this no1 or no2.

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I actually don't need how to explain this. How are you supposed to know whether to write the equation as when x is between two numbers (Like in no1) or X is greater than/less to each number as is No2.

For no. 2 I mean X is less than -3.

Also, in No1 why is X less than 1 and not less than or equal to. I'm confused at to why the inequality changes.
(edited 8 years ago)

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Reply 1
Original post by Questioness

I actually don't need how to explain this. How are you supposed to know whether to write the equation as when x is between two numbers (Like in no1) or X is greater than/less to each number as is No2.

For no. 2 I mean X is less than -3.


By looking at the graph. If the shaded portion is in between two values then you put x as inbeween two values. As you have for Q1, for Q2 the shaded region is all the right of 2 and all the left of -3.
Reply 2
Original post by Zacken
By looking at the graph. If the shaded portion is in between two values then you put x as inbeween two values. As you have for Q1, for Q2 the shaded region is all the right of 2 and all the left of -3.

Only the equation is the given. I drew the graph because I know what to do I just understand when to do it or why.
Reply 3
Original post by Questioness
Slightly confused as to when I do this no1 or no2.

image.jpg
I actually don't need how to explain this. How are you supposed to know whether to write the equation as when x is between two numbers (Like in no1) or X is greater than/less to each number as is No2.

For no. 2 I mean X is less than -3.


Well, you shouldn't have got a quadratic for number one; remember if the fraction is equal to zero, the numerator must be.

For number two you have the right idea, draw the graph and see where the y values are above 0
Reply 4
Original post by Zacken
By looking at the graph. If the shaded portion is in between two values then you put x as inbeween two values. As you have for Q1, for Q2 the shaded region is all the right of 2 and all the left of -3.


I'm a slow typist; but I'm surprised you didn't notice the mistake in Q1 :wink:
Reply 5
Original post by Andy98
Well, you shouldn't have got a quadratic for number one; remember if the fraction is equal to zero, the numerator must be.


Huh? Look at it again.
Reply 6
Original post by Andy98
I'm a slow typist; but I'm surprised you didn't notice the mistake in Q1 :wink:


Because there isn't one.
Reply 7
Original post by Questioness
Only the equation is the given. I drew the graph because I know what to do I just understand when to do it or why.


You're meant to sketch the graph each time.
saw the title, thought it was politics. Mb lol.
Reply 9
Original post by Questioness


Also, in No1 why is X less than 1 and not less than or equal to. I'm confused at to why the inequality changes.


If x=1x=1 then you end up with a division by 0.
Reply 10
Original post by Zacken
Huh? Look at it again.


Original post by Zacken
Because there isn't one.


Well where does the quadratic come from?

Here is what I would do:

x4x1<0\frac{x-4}{x-1}<0 implies x4<0x-4<0, thus x<4x<4
Reply 11
Original post by Andy98
Well where does the quadratic come from?

Here is what I would do:

x4x1<0\frac{x-4}{x-1}<0 implies x4<0x-4<0, thus x<4x<4


Classic error. What if x=10x = -10? 104101=1411=1411>0\frac{-10 - 4}{-10 - 1} = \frac{-14}{-11} = \frac{14}{11} > 0 Yet 10<4-10 < 4

What you've done is you've multiplied both sides of the inequality by x1x-1 which might be negative so that's invalid.

What the OP has done is multiplied both side by (x1)2(x-1)^2 by which is always positive.
Reply 12
Original post by Zacken
By looking at the graph. If the shaded portion is in between two values then you put x as inbeween two values. As you have for Q1, for Q2 the shaded region is all the right of 2 and all the left of -3.


And how will I know where to shade? If I don't know if the equation is 'in between' or not.
Reply 13
Original post by Questioness
And how will I know where to shade? If I don't know if the equation is 'in between' or not.


When your inequality is quadratic > 0: Look at your curve and see where the curve is above or below the x-axis. Shade the regions for which the curve is above the x-axis.

When your inequality is quadratic < 0: Look at your curve and see where the curve is above or below the x-axis. Shade the regions for which the curve is below the x-axis.
Reply 14
Original post by Zacken
When your inequality is quadratic > 0: Look at your curve and see where the curve is above or below the x-axis. Shade the regions for which the curve is above the x-axis.

When your inequality is quadratic < 0: Look at your curve and see where the curve is above or below the x-axis. Shade the regions for which the curve is below the x-axis.

Ohhh right!!!:yeah:
Thanks Zacken, I understand now.
Reply 15
Original post by Questioness
Ohhh right!!!:yeah:
Thanks Zacken, I understand now.


Yay! That clicking feeling in your head is amazing, isn't it? :biggrin:

You're very welcome! :smile:
Reply 16
Original post by Zacken
Yay! That clicking feeling in your head is amazing, isn't it? :biggrin:

You're very welcome! :smile:

Sure is! :tongue:
Reply 17
Original post by Questioness
Sure is! :tongue:


Awesome. :-)
Reply 18
Original post by Zacken
Classic error. What if x=10x = -10? 104101=1411=1411>0\frac{-10 - 4}{-10 - 1} = \frac{-14}{-11} = \frac{14}{11} > 0 Yet 10<4-10 < 4

What you've done is you've multiplied both sides of the inequality by x1x-1 which might be negative so that's invalid.

What the OP has done is multiplied both side by (x1)2(x-1)^2 by which is always positive.


I'm getting sick of you proving my teachers wrong :tongue:
Reply 19
Original post by Andy98
I'm getting sick of you proving my teachers wrong :tongue:


Sorry! At least now you know. :biggrin:

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