Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    0
    ReputationRep:
    This feels so simple and I don't know why I can't do it.

    Express 4x^2 -32x +17 in the form a(x-b)^2 +c

    So I have done
    4(x-8x)+17
    then 4(x-4)^2 -16 +17 (+1)
    I'm not sure if 16 should be positive or negative but I don't see how it works either way because it doesn't go back to what it was when I expand it. What am I doing wrong?
    • PS Helper
    • Study Helper
    Offline

    13
    ReputationRep:
    PS Helper
    Study Helper
    (Original post by jadechapell)
    This feels so simple and I don't know why I can't do it.

    Express 4x^2 -32x +17 in the form a(x-b)^2 +c

    So I have done
    4(x-8x)+17
    then 4(x-4)^2 -16 +17 (+1)
    I'm not sure if 16 should be positive or negative but I don't see how it works either way because it doesn't go back to what it was when I expand it. What am I doing wrong?
    You haven't multiplied the -16 by its 4.

    4x^2-32x+17 = 4(x^2-8x) + 17 = 4[(x-4)^2-16]+4 \times 16 + 17.

    ETA: Typo: the last 4 \times 16 shouldn't be there. That's what happens when I don't reread what I've written.
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by Smaug123)
    You haven't multiplied the -16 by its 4.

    4x^2-32x+17 = 4(x^2-8x) + 17 = 4[(x-4)^2-16]+4 \times 16 + 17.
    Where did
    +4*16+17 come from?
    • PS Helper
    • Study Helper
    Offline

    13
    ReputationRep:
    PS Helper
    Study Helper
    (Original post by jadechapell)
    Where did
    +4*16+17 come from?
    Apologies, typo - I didn't reread. Scratch the 4*16.

    4x^2-32x+17 = 4(x^2-8x) + 17 = 4[(x-4)^2-16] + 17
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by Smaug123)
    Apologies, typo - I didn't reread. Scratch the 4*16.

    4x^2-32x+17 = 4(x^2-8x) + 17 = 4[(x-4)^2-16] + 17
    Ahh okay, so it equals 4(x-4)^2 -47?
    Then if that's right and you were to solve it when it =0 so that the answer is in surd form would you just start off by going 4(x-4)^2 = 47 and then take everything to the other side?
    • PS Helper
    • Study Helper
    Offline

    13
    ReputationRep:
    PS Helper
    Study Helper
    (Original post by jadechapell)
    Ahh okay, so it equals 4(x-4)^2 -47?
    Then if that's right and you were to solve it when it =0 so that the answer is in surd form would you just start off by going 4(x-4)^2 = 47 and then take everything to the other side?
    That's right. So to solve the equation, you'd do (x-4)^2 = \frac{47}{4}, so x-4 = \pm \frac{\sqrt{47}}{2}.
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by Smaug123)
    That's right. So to solve the equation, you'd do (x-4)^2 = \frac{47}{4}, so x-4 = \pm \frac{\sqrt{47}}{2}.
    Thank you!
 
 
 
Poll
Do you agree with the PM's proposal to cut tuition fees for some courses?
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.