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    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)
    Answer is x:-1/4<x<0
    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

    But it's meant to be 0... I'm unsure what I've done wrong. Please help me
    Thank you😀
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    4(x+2)=15-(x+7)
    4x+8=15-x-7

    Then you’re fine
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    (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)
    Answer is x:-1/4<x<0
    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

    But it's meant to be 0... I'm unsure what I've done wrong. Please help me
    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.
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    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
    =
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    (Original post by Pastelx)
    4(x+2)=15-(x+7)
    4x+8=15-x-7

    Then you’re fine
    Yay thanks
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    (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
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    (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
    Extremely helpful
 
 
 

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