Turn on thread page Beta

Another P1 thing watch

    • Thread Starter
    Offline

    14
    ReputationRep:
    m and n are negative constants such that

    (m - √3)(1 + n√3) = 5√3 - 7

    Find the exact values of m and n.

    I tried multiplying out, then realised that was going the wrong way and that they were identities, but I'm used to adding ones. So I put the first bracket equal to 1, and got a value for n, then realised it said "m and n are...constants", and screamed. So I gave multiplying out a go, then realised equating coefficients is again what you do when the letters are variables.

    Help!
    Offline

    2
    ReputationRep:
    (Original post by mik1a)
    m and n are negative constants such that

    (m - √3)(1 + n√3) = 5√3 - 7

    Find the exact values of m and n.

    I tried multiplying out, then realised that was going the wrong way and that they were identities, but I'm used to adding ones. So I put the first bracket equal to 1, and got a value for n, then realised it said "m and n are...constants", and screamed. So I gave multiplying out a go, then realised equating coefficients is again what you do when the letters are variables.

    Help!
    You are going about this the right way.

    I am probably wrong here but I think the answers are:

    m = [(7-√109)/2]-7

    n = (7-√109)/6

    They just don't seem nice answers
    Offline

    2
    ReputationRep:
    Forget those answers. They're utter crap.
    Offline

    2
    ReputationRep:
    (Original post by mik1a)
    m and n are negative constants such that

    (m - √3)(1 + n√3) = 5√3 - 7

    Find the exact values of m and n.

    I tried multiplying out, then realised that was going the wrong way and that they were identities, but I'm used to adding ones. So I put the first bracket equal to 1, and got a value for n, then realised it said "m and n are...constants", and screamed. So I gave multiplying out a go, then realised equating coefficients is again what you do when the letters are variables.

    Help!
    expanding gives m+mnsqrt3-sqrt 3 -3n so clearly we know that
    m-3n=7 and mn-1=5
    so m=7+3n
    n(7+3n)=5
    =>3n^2+7n-5=0
    n=-7+-sqrt(49+4*5*3)/6=(-7+-sqrt109)/6??? and then u can get m?
    i think i have gone wron sumwheer
    Offline

    13
    ReputationRep:
    I tried it, but then got (m- sqrt3)(1-1) = 5 sqrt 3 -7

    and it doesn't work from thereon.
    Offline

    2
    ReputationRep:
    (Original post by mik1a)
    m and n are negative constants such that

    (m - √3)(1 + n√3) = 5√3 - 7

    Find the exact values of m and n.

    I tried multiplying out, then realised that was going the wrong way and that they were identities, but I'm used to adding ones. So I put the first bracket equal to 1, and got a value for n, then realised it said "m and n are...constants", and screamed. So I gave multiplying out a go, then realised equating coefficients is again what you do when the letters are variables.

    Help!
    Ok i think i've managed it.. reasonable sounding answers.

    (m - √3)(1 + n√3) = 5√3 - 7

    m + mn√3 - √3 - n√3(√3) = 5√3 - 7

    (mn√3 - √3) + (m - 3n) = 5√3 - 7


    terms involving √3 equal..

    therefore eqn1) 5√3 = mn√3 - √3

    and eqn 2) -7 = m - 3n

    so using 2) m = 3n - 7

    sub into 1) 6√3 = (3n - 7)n√3

    so 6 = -7n + 3n^2

    solve quadratic.. (3n + 2)(n - 3) = 0

    n = 3 OR n = -2/3 ... we're told its negative so n = -2/3.

    then using 2) m = 3n - 7
    sub in vals of n..
    m = -9 OR m = 2

    so n = -2/3 , m = -9
    Offline

    2
    ReputationRep:
    (Original post by lgs98jonee)
    expanding gives m+mnsqrt3-sqrt 3 -3n so clearly we know that
    m-3n=7 and mn-1=5
    so m=7+3n
    n(7+3n)=5
    =>3n^2+7n-5=0
    n=-7+-sqrt(49+4*5*3)/6=(-7+-sqrt109)/6??? and then u can get m?
    i think i have gone wron sumwheer
    so m=7+(-7+-sqrt109)/6
    =(35+-sqrt109)/6
    Offline

    2
    ReputationRep:
    (Original post by Sahir)
    so n = -2/3 , m = -9
    that's what i got
    Offline

    2
    ReputationRep:
    (Original post by mockel)
    that's what i got
    Sounds good to me.
    Offline

    2
    ReputationRep:
    (Original post by Sahir)
    Ok i think i've managed it.. reasonable sounding answers.

    (m - √3)(1 + n√3) = 5√3 - 7

    m + mn√3 - √3 - n√3(√3) = 5√3 - 7

    (mn√3 - √3) + (m - 3n) = 5√3 - 7


    terms involving √3 equal..

    therefore eqn1) 5√3 = mn√3 - √3

    and eqn 2) -7 = m - 3n

    so using 2) m = 3n - 7

    sub into 1) 6√3 = (3n - 7)n√3

    so 6 = -7n + 3n^2

    solve quadratic.. (3n + 2)(n - 3) = 0

    n = 3 OR n = -2/3 ... we're told its negative so n = -2/3.

    then using 2) m = 3n - 7
    sub in vals of n..
    m = -9 OR m = 2

    so n = -2/3 , m = -9
    yes!
    Offline

    2
    ReputationRep:
    (Original post by Sahir)
    Ok i think i've managed it.. reasonable sounding answers.

    (m - √3)(1 + n√3) = 5√3 - 7

    m + mn√3 - √3 - n√3(√3) = 5√3 - 7

    (mn√3 - √3) + (m - 3n) = 5√3 - 7


    terms involving √3 equal..

    therefore eqn1) 5√3 = mn√3 - √3

    and eqn 2) -7 = m - 3n

    so using 2) m = 3n - 7

    sub into 1) 6√3 = (3n - 7)n√3

    so 6 = -7n + 3n^2

    solve quadratic.. (3n + 2)(n - 3) = 0

    n = 3 OR n = -2/3 ... we're told its negative so n = -2/3.

    then using 2) m = 3n - 7
    sub in vals of n..
    m = -9 OR m = 2

    so n = -2/3 , m = -9
    I've checked that and it is right. Well done.
    Offline

    2
    ReputationRep:
    (Original post by lgs98jonee)
    yes!
    cant believe i didnt get that right i put m=PLUS7PLUS 3n and not MINUS 7...i can believe how dumn i was :-(((
    • Thread Starter
    Offline

    14
    ReputationRep:
    So you need to equate coefficients of √3? I suppose that's fair since it's an irrational number, no other number could be exactly that.
    • Thread Starter
    Offline

    14
    ReputationRep:
    Find the exact square roots of 11+2√30 in surd form.

    I don't know how you take the square root of (a+b), or at least I don't think I do. I know it's not the same as √a+√b at least.
    Offline

    13
    ReputationRep:
    So it is asking you to find the square root of 11 + 2 sqrt30 ?
    Offline

    2
    ReputationRep:
    (Original post by mik1a)
    Find the exact square roots of 11+2√30 in surd form.

    I don't know how you take the square root of (a+b), or at least I don't think I do. I know it's not the same as √a+√b at least.
    how do u get the sqrt sign?
    Offline

    2
    ReputationRep:
    11+2√30 =sqrt(121+120)
    =sqrt(241)
    and sqrt of that =241^0.25
    • Thread Starter
    Offline

    14
    ReputationRep:
    microsoft word lets you use symbols, then I copy and paste. there's probably a alt+number combination for it as well.

    2276: yeah I guess so from what the questions says.
    Offline

    2
    ReputationRep:
    I am afraid you're wrong lgs98jonee

    241^0.25 = 3.94007293....

    √(11+2√30) = 4.68555772....
    Offline

    2
    ReputationRep:
    (Original post by lgs98jonee)
    11+2√30 =sqrt(121+120)
    =sqrt(241)
    and sqrt of that =241^0.25
    √a + √b = √(a+b) is incorrect.
 
 
 
Poll
Do you think parents should charge rent?

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.