Hey there! Sign in to join this conversationNew here? Join for free

I'm stuck...Maths Quadratic inequalities Watch

Announcements
    • Thread Starter
    Offline

    9
    ReputationRep:
    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😀
    Offline

    15
    ReputationRep:
    4(x+2)=15-(x+7)
    4x+8=15-x-7

    Then you’re fine
    Posted on the TSR App. Download from Apple or Google Play
    Online

    7
    ReputationRep:
    (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.
    Offline

    6
    ReputationRep:
    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
    =
    • Thread Starter
    Offline

    9
    ReputationRep:
    (Original post by Pastelx)
    4(x+2)=15-(x+7)
    4x+8=15-x-7

    Then you’re fine
    Yay thanks
    • Thread Starter
    Offline

    9
    ReputationRep:
    (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
    • Thread Starter
    Offline

    9
    ReputationRep:
    (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
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    Will you be richer or poorer than your parents?
    Useful resources

    Make your revision easier

    Maths

    Maths Forum posting guidelines

    Not sure where to post? Read the updated guidelines here

    Equations

    How to use LaTex

    Writing equations the easy way

    Student revising

    Study habits of A* students

    Top tips from students who have already aced their exams

    Study Planner

    Create your own Study Planner

    Never miss a deadline again

    Polling station sign

    Thinking about a maths degree?

    Chat with other maths applicants

    Can you help? Study help unanswered threads

    Groups associated with this forum:

    View associated groups
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

    Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

    Quick reply
    Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.