Hey there! Sign in to join this conversationNew here? Join for free
x Turn on thread page Beta
    • Thread Starter
    Offline

    17
    ReputationRep:
    |2x+3|=2x^2-9 how can I determine the roots?

    Attachment 730072 Graph looks like this
    • Community Assistant
    Online

    20
    ReputationRep:
    Community Assistant
    (Original post by joyoustele)
    |2x+3|=2x^2-9 how can I determine the roots?

    Graph looks like this
    Solve 2x+3=2x^2-9 for x > -\frac{3}{2} and -(2x+3)=2x^2-9 for x < -\frac{3}{2}

    Hint: you can factorise straight away
    Offline

    10
    ReputationRep:
    (Original post by joyoustele)
    |2x+3|=2x^2-9 how can I determine the roots?

    Attachment 730072 Graph looks like this
    You can solve 2x+3=2x^2-9 and -2x-3=2x^2-9 but beware of false solutions.
    Online

    20
    ReputationRep:
    you need to find where the V shape crosses the U shape.
    • Thread Starter
    Offline

    17
    ReputationRep:
    (Original post by BuryMathsTutor)
    You can solve 2x+3=2x^2-9 and -2x-3=2x^2-9 but beware of false solutions.
    If I didnt have the graph, is there another way?
    • Thread Starter
    Offline

    17
    ReputationRep:
    (Original post by the bear)
    you need to find where the V shape crosses the U shape.
    Yh, I know, but the bit that confuses me is.... You know if the gradient of modulus was negative where intersection happens, you put negative infront of the modulus equation, Can you do the same with the x^2 graph?
    • Thread Starter
    Offline

    17
    ReputationRep:
    (Original post by BuryMathsTutor)
    You can solve 2x+3=2x^2-9 and -2x-3=2x^2-9 but beware of false solutions.
    Yes, Thanks a lot
    • Thread Starter
    Offline

    17
    ReputationRep:
    (Original post by RDKGames)
    Solve 2x+3=2x^2-9 for x > -\frac{3}{2} and -(2x+3)=2x^2-9 for x < -\frac{3}{2}

    Hint: you can factorise straight away
    Okay, TY
    • Community Assistant
    Online

    20
    ReputationRep:
    Community Assistant
    (Original post by joyoustele)
    Okay, TY
    The way to do it without the graph is to realise that |2x+3| is actually just made up from two different straight lines, 2x+3 and -(2x+3), where the changeover happens whenever 2x+3=0 hence where the conditions on x come from.

    This means that, by definition, \displaystyle |2x+3| = \begin{cases} 2x+3, \qquad x \geq -\frac{3}{2} \\ -(2x+3), \quad x < -\frac{3}{2}\end{cases}.
    Online

    20
    ReputationRep:
    (Original post by joyoustele)
    Yh, I know, but the bit that confuses me is.... You know if the gradient of modulus was negative where intersection happens, you put negative infront of the modulus equation, Can you do the same with the x^2 graph?
    if the quadratic is inside a modulus then you can do that.
    • Community Assistant
    • Study Helper
    Offline

    20
    ReputationRep:
    Community Assistant
    Study Helper
    (Original post by joyoustele)
    If I didnt have the graph, is there another way?
    You can do these questions without a graph but you should always draw a graph in an exam - they’re usually quadratic/linear functions so easy to draw.

    Doing it without a graph is risky unless you’re very confident with what you’re doing.
    • Thread Starter
    Offline

    17
    ReputationRep:
    (Original post by Notnek)
    You can do these questions without a graph but you should always draw a graph in an exam - they’re usually quadratic/linear functions so easy to draw.

    Doing it without a graph is risky unless you’re very confident with what you’re doing.
    Okay.
    Thanks
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: March 9, 2018
Poll
Do you agree with the proposed ban on plastic straws and cotton buds?
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

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

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