Results are out! Find what you need...fast. Get quick advice or join the chat
Hey! Sign in to get help with your study questionsNew here? Join for free to post

Additional Maths

Announcements Posted on
Applying to uni this year? Check out our new personal statement advice hub 28-11-2014
    • Thread Starter
    • 3 followers
    Offline

    ReputationRep:
    The cubic equation x^3 + ax^2 + bx -26 = 0 has 3 positive, distinct, integer roots. Find the values a and b.

    I don't understand how to work this out.
    • 1 follower
    Offline

    ReputationRep:
    Have you tried getting everything factorised

    \displaystyle x^3 + ax^2 + bx - 26 =

    \displaystyle p(x) = x^3 - 2x^2 - x + 2 = (x - 1)(x^2 + ax + b) = x^3 + (a - 1)x^2 + (b - a)x - b

    Is that what you mean?
    • Thread Starter
    • 3 followers
    Offline

    ReputationRep:
    What factor would I use?
    • 1 follower
    Offline

    ReputationRep:
    If the equation has three distinct roots, you will be able to factorise it into something of the form:

    \displaystyle (x-f)(x-g)(x-h) = 0

    Where f, g and h are your solutions. When you multiply this all out, you're going to get a bunch of different terms. Importantly, you will find that you have only one constant term, -fgh. You know that this must equal -26, which tells you what your three solutions must be (here you can apply the restriction that the solutions are distinct and positive integers). Put your solutions back into the factorised form and multiply out in full.
    • Thread Starter
    • 3 followers
    Offline

    ReputationRep:
    (Original post by TheMagicMan)
    If r_i are the roots of the polynomial \displaystyle\sum_{i=0}^n a_i x^i, then \prod r_i =(-1)^n a_0. Use the fact that r_i \not = r_j if i \not= j, and that r_i \in N
    I wish I understood this.
    • Thread Starter
    • 3 followers
    Offline

    ReputationRep:
    (Original post by porkstein)
    If the equation has three distinct roots, you will be able to factorise it into something of the form:

    \displaystyle (x-f)(x-g)(x-h) = 0

    Where f, g and h are your solutions. When you multiply this all out, you're going to get a bunch of different terms. Importantly, you will find that you have only one constant term, -fgh. You know that this must equal -26, which tells you what your three solutions must be (here you can apply the restriction that the solutions are distinct and positive integers). Put your solutions back into the factorised form and multiply out in full.
    Thanks.
    • Thread Starter
    • 3 followers
    Offline

    ReputationRep:
    I got 1,2,13, can't it be -1,-2, -13?
    • 3 followers
    Offline

    ReputationRep:
    (Original post by Math12345)
    I got 1,2,13, can't it be -1,-2, -13?
    Notice the condition upon the question - they are three POSITIVE, distinct integer roots.*

    If -1, -2, and -13 are the roots, then you'll have the cubic equation:

     (x+1)(x+2)(x+13)

    But  1 \times  2 \times  13 \neq -26 , so -1, -2 and -13 can't be the three roots (plus they're not even positive in the first place)
    • Thread Starter
    • 3 followers
    Offline

    ReputationRep:
    (Original post by Blazy)
    Notice the condition upon the question - they are three POSITIVE, distinct integer roots.*

    If -1, -2, and -13 are the roots, then you'll have the cubic equation:

     (x+1)(x+2)(x+13)

    But  1 \times  2 \times  13 \neq -26 , so -1, -2 and -13 can't be the three roots (plus they're not even positive in the first place)
    Got it, thanks.
    • 1 follower
    Offline

    ReputationRep:
    (Original post by Math12345)
    If r_i are the roots of the polynomial \displaystyle\sum_{i=0}^n a_i x^i, then \displaystyle \prod r_i =(-1)^n a_0. Use the fact that r_i \not = r_j if i \not= j, and that r_i \in N
    I wish I understood this.
    TheMagicMan is being almost deliberately confusing, introducing notation that simply isn't necessary for this question. He's essentially just expressing what the question says in words with symbols, presumably because that satisfies him or something. I can't see how he could've thought it would help you in any way.

    It really isn't relevant, but in case you'd like to know what he means (it's not as hard as it looks):

    \displaystyle\sum_{i=0}^n a_i x^i

    Means:

    \displaystyle a_0 x^0 + a_1 x + a_2 x^2 + a_3 x^3 + \cdots + a_n x^n

    a4 just represents some general constant that is multiplied by the x4 term. You could call it "the x4 coefficient". As such the expression above involving that capital sigma (the big E-like symbol) is just a way of expressing some general polynomial (of degree n).

    \displaystyle \prod r_i , which should really be written as \displaystyle \prod_{i=1}^n r_i

    Means:

    \displaystyle r_1 \times r_2 \times r_3 \times  \cdots \times r_n

    So looking back at the equation involving this capital pi, what we're trying to say is that the product of all the roots of the equation is equal to a0, the constant term of the polynomial. This is similar to what I said about fgh being equal to 26. The (-1)^n bit just comes about because if, say, 5 is a root of the equation, then in the factorised form it's going to appear as (x minus 5). When all these roots are multiplied together you'll potentially get a minus term floating around, depending on just how many things you multiply together.

    r_i \not = r_j if i \not= j

    This is just a fancy way of saying that the roots are all different. The second root of the equation, r2, is not equal to the fifth root of the equation, r5. Nor is the fifth root equal to the fourth root, and so on. The i just stands in for any number, as does j. When you run through all the combinations of roots, if you ask yourself, say, whether the third root of the equation equals the third root of the equation, the answer is obviously yes. Thus the i =/= j just says that the roots are all different, providing you don't consider the same root twice in your comparisons.

    r_i \in N

    This is just a fancy way of saying the roots are integers.
    • 4 followers
    Offline

    ReputationRep:
    (Original post by porkstein)
    TheMagicMan is being almost deliberately confusing, introducing notation that simply isn't necessary for this question. He's essentially just expressing what the question says in words with symbols, presumably because that satisfies him or something. I can't see how he could've thought it would help you in any way.

    It really isn't relevant, but in case you'd like to know what he means (it's not as hard as it looks):

    \displaystyle\sum_{i=0}^n a_i x^i

    Means:

    \displaystyle a_0 x^0 + a_1 x + a_2 x^2 + a_3 x^3 + \cdots + a_n x^n

    a4 just represents some general constant that is multiplied by the x4 term. You could call it "the x4 coefficient". As such the expression above involving that capital sigma (the big E-like symbol) is just a way of expressing some general polynomial (of degree n).

    \displaystyle \prod r_i , which should really be written as \displaystyle \prod_{i=1}^n r_i

    Means:

    \displaystyle r_1 \times r_2 \times r_3 \times  \cdots \times r_n

    So looking back at the equation involving this capital pi, what we're trying to say is that the product of all the roots of the equation is equal to a0, the constant term of the polynomial. This is similar to what I said about fgh being equal to 26. The (-1)^n bit just comes about because if, say, 5 is a root of the equation, then in the factorised form it's going to appear as (x minus 5). When all these roots are multiplied together you'll potentially get a minus term floating around, depending on how many things you multiply together.

    r_i \not = r_j if i \not= j

    This is just a fancy way of saying that the roots are all different. The second root of the equation, r2, is not equal to the fifth root of the equation, r5. Nor is the fifth root equal to the fourth root, and so on. The i just stands in for any number, as does j. When you run through all the combinations of roots, if you ask yourself, say, whether the third root of the equation equals the third root of the equation, the answer is obviously yes. Thus the i =/= j just says that the roots are all different, providing you don't consider the same root twice in your comparisons.

    r_i \in N

    This is just a fancy way of saying the roots are integers.
    There's really only one hint that's helpful in this problem...that the product of the roots is equal to the coefficient of x^0...as for the notation, it's a lot quicker than words, and most users of this site would understand it.

    Also, when there is an 'obvious' range for the summation and product limits, there is really no need to put them on.

    Anyhow, like most people here, I'm just trying to help, so...
    • 1 follower
    Offline

    ReputationRep:
    (Original post by TheMagicMan)
    ...
    It's certainly not quicker than words - I highly doubt it took you longer to write that 'helpful hint' than the mess of LaTeX you gave him. Given the OP is asking a question about additional maths, which even if one didn't know was a qualification for 16 year-olds is (from the question) evidently not particularly advanced stuff, I find it highly unlikely that (s)he would be comfortable throwing around summations and products and terms with subscript i in that fashion.

    And if you're going to leave the limits off, at least be consistent in doing so.
    • 1 follower
    Offline

    ReputationRep:
    Can somebody please post a solution with the answers

    edit: nevermind
    • 12 followers
    Offline

    ReputationRep:
    I am sorry but why is everone confusing this kid on such a basic question.

    If it is a cubic and the constant is -26, then the factors have to make -26.

    Now let's take all the factors of -26; they are -1,-2,-13,-26.

    A factor can not be -26, since then you would have -1 and -1 as the other two factors which aren't distinct. Therefore, the three factors are -1,-2,-13.

    Therefore we see the equation can be written in this format:

    (x-1)(x-2)(x-13)=0

    The roots of the above equation are x=1,2,13: Three positive, distinct, integer roots!

    Now simply expand the brackets and you get:

    x^3–16x^2+41x–26 = 0.

    Therefore, a =-16 and b=41
    • 12 followers
    Offline

    ReputationRep:
    (Original post by TheMagicMan)
    If r_i are the roots of the polynomial \displaystyle\sum_{i=0}^n a_i x^i, then \prod r_i =(-1)^n a_0. Use the fact that r_i \not = r_j if i \not= j, and that r_i \in N
    Don't try and act really clever with this. You are being annoying by trying to look so great and confusing the kid, when there is a very easy solution which you have failed to find with your "complex" method.
    • 12 followers
    Offline

    ReputationRep:
    (Original post by Blazy)
    Notice the condition upon the question - they are three POSITIVE, distinct integer roots.*

    If -1, -2, and -13 are the roots, then you'll have the cubic equation:

     (x+1)(x+2)(x+13)

    But  1 \times  2 \times  13 \neq -26 , so -1, -2 and -13 can't be the three roots (plus they're not even positive in the first place)
    i think he is confused between factors and roots.
    • 4 followers
    Offline

    ReputationRep:
    (Original post by GreenLantern1)
    Don't try and act really clever with this. You are being annoying by trying to look so great and confusing the kid, when there is a very easy solution which you have failed to find with your "complex" method.
    It's not complex...it's exactly the same as whatever else has been suggested
    • 12 followers
    Offline

    ReputationRep:
    (Original post by TheMagicMan)
    It's not complex...it's exactly the same as whatever else has been suggested
    No you have suggested a complex notation that the kid is clearly not going to understand when he is only 15/16 years old an is doing maths harder than the GCSE normally taken at his age.
    • 15 followers
    Online

    ReputationRep:
    (Original post by TheMagicMan)
    It's not complex...it's exactly the same as whatever else has been suggested
    You know what he means. There was just no need to use higher level notation for this question.

    When you help someone with maths (in TSR or anywhere else), you should first consider the level and ability of the person that you are helping. Sometimes, you may think that you're helping out but really, you're just making them more confused.
    • 12 followers
    Offline

    ReputationRep:
    (Original post by TheMagicMan)
    If r_i are the roots of the polynomial \displaystyle\sum_{i=0}^n a_i x^i, then a_n \prod r_i =(-1)^n a_0. Use the fact that r_i \not = r_j if i \not= j, and that r_i \in N
    Did it make you feel better to give me a negative rating when you clearly were being an arrogant, pompous fool!

Reply

Submit reply

Register

Thanks for posting! You just need to create an account in order to submit the post
  1. this can't be left blank
    that username has been taken, please choose another Forgotten your password?
  2. this can't be left blank
    this email is already registered. Forgotten your password?
  3. this can't be left blank

    6 characters or longer with both numbers and letters is safer

  4. this can't be left empty
    your full birthday is required
  1. By joining you agree to our Ts and Cs, privacy policy and site rules

  2. Slide to join now Processing…

Updated: April 1, 2012
New on TSR

Vote for your favourite Christmas film

Win a bundle of Xmas DVDs

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