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# I'm stuck...Maths Quadratic inequalities watch

1. Use set notation to describe the set of values of x for which:
4x^2-3x-1<0 and 4(x+2)<15-(x+7)
So far I have:
4x(x+1)+(x-1)<0
4x+1=0
4x=-1
X=-1/4

X-1=0
X=1

4(x+2) = 15-(x+7)
4x+8= 15-x+7
5x= 15+7-8
X=14/5

Thank you😀
2. 4(x+2)=15-(x+7)
4x+8=15-x-7

Then you’re fine
3. (Original post by Musicanor)
Use set notation to describe the set of values of x for which:
4x^2-3x-1<0 and 4(x+2)<15-(x+7)
So far I have:
4x(x+1)+(x-1)<0
4x+1=0
4x=-1
X=-1/4

X-1=0
X=1

4(x+2) = 15-(x+7)
4x+8= 15-x+7
5x= 15+7-8
X=14/5

Thank you😀
EDIT: Ignore this bit - I found the solution, sorry [For the quadratic, you should have two solutions, of which you only appear to have one. Check that you factorised correctly. Then sketch a graph to check the signs]

Your mistake on the second one is very easy to make. You've said that 15 - (x+7) = 15 - x + 7 when in reality, the - applies to both
the x and the 7, if that makes sense.
4. basically you have to come up with an inequality that satisfies both inequalities

1st equation you sort of got it , you just had to use the inequality signs not = sign
...
...
(4x+1)(x-1)<0
(after sketching out graph i found --> -1/4<x<1

second equation you did a mistake --> 4(x+2)<15-(x+7)--> you didnt open the bracket properly ... 7 should be minus as there is a minus before the whole bracket

--> 4x+8<15-x-7
-->5x<0
--> x<0

now because you have to come up with an inequality that satisfies both you have to see where the 2 coincides .. you can draw out a number line using inequality symbols etc.

thats how you come up with the innequality -1/4<x<0
=
5. (Original post by Pastelx)
4(x+2)=15-(x+7)
4x+8=15-x-7

Then you’re fine
Yay thanks
6. (Original post by ecila21)
EDIT: Ignore this bit - I found the solution, sorry [For the quadratic, you should have two solutions, of which you only appear to have one. Check that you factorised correctly. Then sketch a graph to check the signs]

Your mistake on the second one is very easy to make. You've said that 15 - (x+7) = 15 - x + 7 when in reality, the - applies to both
the x and the 7, if that makes sense.
Thanks
7. (Original post by noorah.dj)
basically you have to come up with an inequality that satisfies both inequalities

1st equation you sort of got it , you just had to use the inequality signs not = sign
...
...
(4x+1)(x-1)<0
(after sketching out graph i found --> -1/4<x<1

second equation you did a mistake --> 4(x+2)<15-(x+7)--> you didnt open the bracket properly ... 7 should be minus as there is a minus before the whole bracket

--> 4x+8<15-x-7
-->5x<0
--> x<0

now because you have to come up with an inequality that satisfies both you have to see where the 2 coincides .. you can draw out a number line using inequality symbols etc.

thats how you come up with the innequality -1/4<x<0
=
Thank you

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