Notices
Results 1 to 37 of 37

Thread: Maths homework (solving quadratics by factoring)

  1. #1 Maths homework (solving quadratics by factoring) 
    Forum Sophomore somfooleishfool's Avatar
    Join Date
    May 2011
    Location
    New Zealand
    Posts
    196
    I'm a bit stuck.

    I've just learnt this today and I'm just working throught the textbook learning as I go.

    in examples like (x+3)(x-7)=0 I have no trouble working out that x = -3 or 7

    I also have no trouble with x^2+5x-36=0 x= -9 or 4

    But as I get through the equations they are adding things in bit by bit and now I'm stumped by something that is quite possibly very simple but there is no example showing how to do it.

    HERE is what I'm stuck on
    2x^2-15x+28=0

    I can't work out what to do with 2x^2 as opposed to the 1x^2 I've used so far.


    Reply With Quote  
     

  2.  
     

  3. #2  
    Forum Ph.D. Heinsbergrelatz's Avatar
    Join Date
    Aug 2009
    Location
    Singapore
    Posts
    994
    there is a technique required known as 'completing the square'. Its pretty neat, and its used often when the a is greater than 1, but not always.

    but before you solve anything check if b^2-4ac>0 first.

    anyway this is what you need to do. you need to make the equation in a form.
    a(x-h)^2+c=0

    divide 2 to all sides and you get;

    x^2-7.5x+14=0
    then you solve it as if you would solve a quadratic with ax^2+... where a=1.
    so you would get;

    (x-3.75)^2+c=0

    now you need to find this c, as clearly 3.75^2 is not 14. so c in this case is -1/16.
    so it would like,

    (x-3.75)^2-1/16=0

    or you could use the normal factorization method. Note completing the square if useful is factorization cannot take place.

    which gives, (2x-7)(x-4)=0


    ------------------




    "Mathematicians stand on each other's shoulders."- Carl Friedrich Gauss


    -------------------
    Reply With Quote  
     

  4. #3  
    Forum Sophomore somfooleishfool's Avatar
    Join Date
    May 2011
    Location
    New Zealand
    Posts
    196
    Quote Originally Posted by Heinsbergrelatz
    there is a technique required known as 'completing the square'. Its pretty neat, and its used often when the a is greater than 1, but not always.

    but before you solve anything check if b^2-4ac>0 first.

    anyway this is what you need to do. you need to make the equation in a form.
    a(x-h)^2+c=0

    divide 2 to all sides and you get;

    x^2-7.5x+14=0
    then you solve it as if you would solve a quadratic with ax^2+... where a=1.
    so you would get;

    (x-3.75)^2+c=0

    now you need to find this c, as clearly 3.75^2 is not 14. so c in this case is -1/16.
    so it would like,

    (x-3.75)^2-1/16=0

    or you could use the normal factorization method. Note completing the square if useful is factorization cannot take place.

    which gives, (2x-7)(x-4)=0
    Although your probably right and I'm not disputing that, I have several issues with this

    a) I don't know what "h" is ***"anyway this is what you need to do. you need to make the equation in a form.
    a(x-h)^2+c=0"***

    b) The equation I posted "2x^2-15x+28=0" is in my text book under excersises for "solving by factorising" so regardless of whether it CAN be done your way, Id like to know how to do it the way the text book is asking.

    c) We got taught "completing the square" method yesterday and today and so far I don't understand it in the slightest but the method you are using does NOT look like the method we are using at university and as far as I'm aware they will want to see in tests that I worked it out the way they taught us.

    d) Regardless, I still really appreciate you taking the time to help me out on this. Thank you

    Oh and the method of "completing the square we got taught goes a little like this (makes no sence to me but here it is)

    x= !!-b + or - √b^2-4ac!! and I'm not sure how to express this but everything between "!!" is a numerater(!!) to the denominator(&&) 2a

    next step is:

    =!!-4+or-√4^2-4*1*-7!! &&2*1&&

    =!!-4+or-√44!! &&2&&

    =1.317 or -5.317 (both to 4sf)
    Reply With Quote  
     

  5. #4  
    Forum Ph.D. Heinsbergrelatz's Avatar
    Join Date
    Aug 2009
    Location
    Singapore
    Posts
    994
    Quote Originally Posted by somfooleishfool
    Quote Originally Posted by Heinsbergrelatz
    there is a technique required known as 'completing the square'. Its pretty neat, and its used often when the a is greater than 1, but not always.

    but before you solve anything check if b^2-4ac>0 first.

    anyway this is what you need to do. you need to make the equation in a form.
    a(x-h)^2+c=0

    divide 2 to all sides and you get;

    x^2-7.5x+14=0
    then you solve it as if you would solve a quadratic with ax^2+... where a=1.
    so you would get;

    (x-3.75)^2+c=0

    now you need to find this c, as clearly 3.75^2 is not 14. so c in this case is -1/16.
    so it would like,

    (x-3.75)^2-1/16=0

    or you could use the normal factorization method. Note completing the square if useful is factorization cannot take place.

    which gives, (2x-7)(x-4)=0
    Although your probably right and I'm not disputing that, I have several issues with this

    a) I don't know what "h" is ***"anyway this is what you need to do. you need to make the equation in a form.
    a(x-h)^2+c=0"***

    b) The equation I posted "2x^2-15x+28=0" is in my text book under excersises for "solving by factorising" so regardless of whether it CAN be done your way, Id like to know how to do it the way the text book is asking.

    c) We got taught "completing the square" method yesterday and today and so far I don't understand it in the slightest but the method you are using does NOT look like the method we are using at university and as far as I'm aware they will want to see in tests that I worked it out the way they taught us.

    d) Regardless, I still really appreciate you taking the time to help me out on this. Thank you

    Oh and the method of "completing the square we got taught goes a little like this (makes no sence to me but here it is)

    x= !!-b + or - √b^2-4ac!! and I'm not sure how to express this but everything between "!!" is a numerater(!!) to the denominator(&&) 2a

    next step is:

    =!!-4+or-√4^2-4*1*-7!! &&2*1&&

    =!!-4+or-√44!! &&2&&

    =1.317 or -5.317 (both to 4sf)
    you still do this stuff in Uni.? well i guess if you just started uni. maths well the professors maybe wanna go over the basics, or if taking linear algebra but i dont know.

    anyway the general completing the square method that applies to ALL quadratic equations is in the form of a(x-h)^2 +c
    if you dont understand it, i will derive the whole thing, to make it more simple.






    it follows that f(x)>0 for all iff a>0, and D<0.

    now ihere is the factoring part.

    if , then;







    where alpha and beta are your answers.
    ------------------




    "Mathematicians stand on each other's shoulders."- Carl Friedrich Gauss


    -------------------
    Reply With Quote  
     

  6. #5  
    Veracity Vigilante inow's Avatar
    Join Date
    Oct 2009
    Location
    Austin, TX
    Posts
    3,500
    Quote Originally Posted by Heinsbergrelatz
    you still do this stuff in Uni.? well i guess if you just started uni. maths well the professors maybe wanna go over the basics, or if taking linear algebra but i dont know.
    The education system in the US is not quite as robust as it is in Spore.
    Reply With Quote  
     

  7. #6  
    Forum Bachelors Degree x(x-y)'s Avatar
    Join Date
    Jul 2010
    Location
    UK
    Posts
    462
    Quote Originally Posted by inow
    Quote Originally Posted by Heinsbergrelatz
    you still do this stuff in Uni.? well i guess if you just started uni. maths well the professors maybe wanna go over the basics, or if taking linear algebra but i dont know.
    The education system in the US is not quite as robust as it is in Spore.
    Maybe, but "somefoolishfool" is not in the US- he is in New Zealand! :-D

    Anyway, "Heisenbergrelatz" explained completing the square well- but seeing as you want to see how your problem of



    factorises, then I'll show you:



    the case here is to just think of factors of the c value- which, in this case, is 28 (so, 28 and 1, 14 and 2, 7 and 4)- then look at the b value (which is -15) and using the a value (2) try to think logically about which values will work, so this will obviously turn out to be



    Remember that -7 multiplied by -4 will give +28.
    "Nature doesn't care what we call it, she just does it anyway" - R. Feynman
    Reply With Quote  
     

  8. #7  
    Veracity Vigilante inow's Avatar
    Join Date
    Oct 2009
    Location
    Austin, TX
    Posts
    3,500
    Quote Originally Posted by x(x-y)
    Maybe, but "somefoolishfool" is not in the US- he is in New Zealand!
    DOH!!!!!


    Never mind me. I was just passing through.
    Reply With Quote  
     

  9. #8  
    Your Mama! GiantEvil's Avatar
    Join Date
    Apr 2010
    Location
    Vancouver, Wa
    Posts
    1,919
    Quote Originally Posted by somfooleishfool
    I can't work out what to do with 2x^2 as opposed to the 1x^2 I've used so far.
    2x^2.
    Let the coefficient "ride with" the variable when you factor out the equation.
    For example, 4x^2-blah+blah, might factor out as (4x+blah)(x-blah), or as (2x+blah)(2x-blah).
    I was some of the mud that got to sit up and look around.
    Lucky me. Lucky mud.
    -Kurt Vonnegut Jr.-
    Cat's Cradle.
    Reply With Quote  
     

  10. #9  
    Forum Freshman
    Join Date
    Dec 2010
    Posts
    78
    Som-F-F:
    Back in the day it took me quite a while to be able to "see" the relationships when factoring.

    Like this one: two things have to multiply to get 28.

    Then, those two things have to add to get -15 but one of them is multiplied by 2 first.

    So if you work with the possibilities you see that 4 and 7 get your 28; and then if the 4 is multilpied by 2 it adds to 7 to get 15.

    Of course, since the 28 is positive and the 15 is negative it has to be the work of two negative numbers (-4 and -7).

    With practice the process becomes easier.
    Reply With Quote  
     

  11. #10  
    Forum Sophomore somfooleishfool's Avatar
    Join Date
    May 2011
    Location
    New Zealand
    Posts
    196
    I really ought to have mentioned at the start my lack of math education.

    I failed NCEA level 1 then dropped out. I don't know if this means anything to you people in other countries but that is the earliest and lowest form of education available.

    I am now 20 (allowing me entry into university) and since school I have matured a bit and started taking studys seriously so now I am doing what they call a "bridging course" to bring my education up to a university level before I start a bachelor of science.

    What this means to you guys is that anything I did know about maths, was learnt 4 years ago and has since been forgotten, and now my maths knowledge consists of 2 weeks of university study. Aside from this my highest knowledge of maths would be not far beyond my times tables.

    so basically zero symbols if there unexplained please. If you asked me 2 days ago, I wouldn't have recognised the significance of ax^2+bx+c. It meant nothing to me. Equally in Heinsbergrelatz post f=x etc etc. I dont know what "f" means. I don't know what a "coefficaint" is etc etc. Now I'm sure you understand why I'm not picking this up strait away
    Reply With Quote  
     

  12. #11  
    Forum Sophomore somfooleishfool's Avatar
    Join Date
    May 2011
    Location
    New Zealand
    Posts
    196
    Quote Originally Posted by x(x-y)
    Quote Originally Posted by inow
    Quote Originally Posted by Heinsbergrelatz
    you still do this stuff in Uni.? well i guess if you just started uni. maths well the professors maybe wanna go over the basics, or if taking linear algebra but i dont know.
    The education system in the US is not quite as robust as it is in Spore.
    Maybe, but "somefoolishfool" is not in the US- he is in New Zealand! :-D

    Anyway, "Heisenbergrelatz" explained completing the square well- but seeing as you want to see how your problem of



    factorises, then I'll show you:



    the case here is to just think of factors of the c value- which, in this case, is 28 (so, 28 and 1, 14 and 2, 7 and 4)- then look at the b value (which is -15) and using the a value (2) try to think logically about which values will work, so this will obviously turn out to be



    Remember that -7 multiplied by -4 will give +28.
    :O You were doing such a good job explaining, then all of a sudden BOOM "logically here's the answer."

    start again please from here? "and using the a value (2) try to think logically about which values will work, so this will obviously turn out to be etc"

    Thank you all very much for your help and I hope I haven't infuriated too many of you with my 1337 smarts.

    edit: It occurs to me, based on responces, that I haven't explained very well where it is that I'm stuck. I simply don't know what to do with 2a^2 I have only been taught so far what to do with 1a^2. I simply don't know the process to use.
    Reply With Quote  
     

  13. #12  
    Your Mama! GiantEvil's Avatar
    Join Date
    Apr 2010
    Location
    Vancouver, Wa
    Posts
    1,919
    Quote Originally Posted by somfooleishfool
    I can't work out what to do with 2x^2 as opposed to the 1x^2 I've used so far.
    2x^2
    Let the coefficient (the red part) "ride with" the variable (the blue part) when you factor out the equation.
    For example, 4x^2-blah+blah, might factor out as (4x+blah)(x-blah), or as (2x+blah)(2x-blah).
    The answer is (2x-7)(x-4)=0.
    I was some of the mud that got to sit up and look around.
    Lucky me. Lucky mud.
    -Kurt Vonnegut Jr.-
    Cat's Cradle.
    Reply With Quote  
     

  14. #13  
    Forum Ph.D. Heinsbergrelatz's Avatar
    Join Date
    Aug 2009
    Location
    Singapore
    Posts
    994
    Quote Originally Posted by somfooleishfool
    I really ought to have mentioned at the start my lack of math education.

    I failed NCEA level 1 then dropped out. I don't know if this means anything to you people in other countries but that is the earliest and lowest form of education available.

    I am now 20 (allowing me entry into university) and since school I have matured a bit and started taking studys seriously so now I am doing what they call a "bridging course" to bring my education up to a university level before I start a bachelor of science.

    What this means to you guys is that anything I did know about maths, was learnt 4 years ago and has since been forgotten, and now my maths knowledge consists of 2 weeks of university study. Aside from this my highest knowledge of maths would be not far beyond my times tables.

    so basically zero symbols if there unexplained please. If you asked me 2 days ago, I wouldn't have recognised the significance of ax^2+bx+c. It meant nothing to me. Equally in Heinsbergrelatz post f=x etc etc. I dont know what "f" means. I don't know what a "coefficaint" is etc etc. Now I'm sure you understand why I'm not picking this up strait away
    i actually know new zealand pretty well, i once lived there and attended St.Kentigern in Auckland. Alot of my friends were taking NCEA bt i took A-levels. im not trying to offend you, but i did realize alot of the students in new zealand lacked in mathematics, i dont know why but they had problems with their times table even.

    anyway f(x) just means a function, you will come across soon or later, its not that hard, if you understand the definition rigorously enough. The a,b,c are simple coefficients which represent just numbers or symbols, you can almost think of them as a constant.

    Anyway which Uni, are you in? University of auckland?
    ------------------




    "Mathematicians stand on each other's shoulders."- Carl Friedrich Gauss


    -------------------
    Reply With Quote  
     

  15. #14  
    Forum Bachelors Degree x(x-y)'s Avatar
    Join Date
    Jul 2010
    Location
    UK
    Posts
    462
    Quote Originally Posted by somfooleishfool
    so basically zero symbols if there unexplained please. If you asked me 2 days ago, I wouldn't have recognised the significance of ax^2+bx+c. It meant nothing to me. Equally in Heinsbergrelatz post f=x etc etc. I dont know what "f" means. I don't know what a "coefficaint" is etc etc. Now I'm sure you understand why I'm not picking this up strait away
    The thing to do is not worry about symbols which may look horribly complicated- just keep a logical mind about it and think about why symbols may be where they are or what they actually mean.

    So, f(x) just means a function- for example:

    A function could be



    which just means that the function of x is x squared. So,



    this just means that when plugging 2 into this function of x, you will arrive at an answer of 4. It can be more understandable by thinking of f(x) as the variable y- as in the y-axis plot.

    Anyway, you'll probably learn this later on.

    A coefficient is just the number before the unknown variable- for example:

    the coefficient of x squared in



    is 5. It is just the number before the variable.

    Quote Originally Posted by somefoolishfool
    It occurs to me, based on responces, that I haven't explained very well where it is that I'm stuck. I simply don't know what to do with 2a^2 I have only been taught so far what to do with 1a^2. I simply don't know the process to use.
    The process is, essentially, exactly the same as using any other coefficient of the x squared variable. The only difference is that the number in front of the x squared (the coefficient of x squared) is 2 instead of one, and so it is required that in your brackets (for factorising) you will need:



    where , is just a plus or minus symbol depending upon the expression to be factorised and "p" and "q" are just numbers which will multiply to obtain the value of "c" in .

    Hope that helped!
    "Nature doesn't care what we call it, she just does it anyway" - R. Feynman
    Reply With Quote  
     

  16. #15  
    Forum Sophomore somfooleishfool's Avatar
    Join Date
    May 2011
    Location
    New Zealand
    Posts
    196
    Thanks guys! I really appreciate your patience and willingness to help. I get it now

    edit: "which uni are you at" I'm at Canterbury uni
    Reply With Quote  
     

  17. #16  
    Forum Sophomore somfooleishfool's Avatar
    Join Date
    May 2011
    Location
    New Zealand
    Posts
    196
    In example 20x^2 -44x-15=0

    20x^2*-15=-300x^2

    so then how do I know if its
    (20x-50)(x+6)
    or
    (20x+6)(x-50)
    Reply With Quote  
     

  18. #17  
    . DrRocket's Avatar
    Join Date
    Aug 2008
    Posts
    5,486
    Quote Originally Posted by somfooleishfool
    In example 20x^2 -44x-15=0

    20x^2*-15=-300x^2

    so then how do I know if its
    (20x-50)(x+6)
    or
    (20x+6)(x-50)
    20x^2 -44x-15=0 and 20x^2*-15=-300x^2 have no obvious relationship to one another and neither is (20x-50)(x+6) or (20x+6)(x-50).

    Maybe you should consider a different major.

    By inspection, 20x^2 -44x-15= (10x +3)(2x-5)
    Reply With Quote  
     

  19. #18  
    Forum Sophomore somfooleishfool's Avatar
    Join Date
    May 2011
    Location
    New Zealand
    Posts
    196
    Quote Originally Posted by DrRocket
    Quote Originally Posted by somfooleishfool
    In example 20x^2 -44x-15=0

    20x^2*-15=-300x^2

    so then how do I know if its
    (20x-50)(x+6)
    or
    (20x+6)(x-50)
    20x^2 -44x-15=0 and 20x^2*-15=-300x^2 have no obvious relationship to one another and neither is (20x-50)(x+6) or (20x+6)(x-50).

    Maybe you should consider a different major.

    By inspection, 20x^2 -44x-15= (10x +3)(2x-5)
    Thank you.

    I realized this about an hour after posting. Just having trouble teaching myself this maths as I'm not picking up much in lectures. I'm likely to fail my 1st assignment this tuesday as I'm a bit behind.

    I find the textbook to be lacking in assumed knowledge. I found there is a whole other step needed in quadratics when a is not = 1 in ax^2. This for some reason was not explained in the textbook but I found it in my lecture notes.

    "Maybe you should consider a different major."

    I'm taking my degree at half speed. half the papers, double the total degree time. I don't care how long it takes me to learn this. It's my biggest and only real passion.
    once I get a math tutor sorted out I should be fine.
    If I fail. Ill try again.
    What IS annoying is the amount of study time I put in to be falling behind
    Reply With Quote  
     

  20. #19  
    Your Mama! GiantEvil's Avatar
    Join Date
    Apr 2010
    Location
    Vancouver, Wa
    Posts
    1,919
    Here goes...

    Let's start with the equation 20x^2 -44x -15 =0, and were just going to forget about the =0 part for now.

    The part were going to concern ourselves with, 20x^2 -44x -15, is referred to as a POLYNOMIAL.
    POLY means "many" or "several", "more than one". NOMIAL means "number". So polynomial means "many numbers".

    Each part of the polynomial is called a TERM. The parts or term's are [20x^2], and [-44x], and [-15].

    Each term has a COEFFICIENT which is the number part, [20x^2], [-44x], [-15]
    Each term may or may not have a VARIABLE. The variable is the letter part, [20x^2], [-44x].
    If a term has a coefficient but no variable, then that term is called a CONSTANT.
    In our particular polynomial, [-15] is a constant.

    Taking this, 20x^2-44x-15, and turning it into this, (10x+3)(2x-5), is called FACTORING.
    Factoring is when a term or polynomial is broken down into smaller parts, but only under the operation of multiplication.
    By the way, this (10x+3), and this (2x-5), are each called a BINOMIAL.
    BI means "two", and you know what "nomial" means.

    To turn our binomials, (10x+3)(2x-5), back into our polynomial, 20x^2-44x-15, we can use the F.O.I.L. method.
    FIRST we recombine the first terms(10x+3)(2x-5) for 20x^2.
    OUTSIDE terms are recombined(10x+3)(2x-5) for 20x^2-50x.
    INSIDE terms (10x+3)(2x-5) for 20x^2-50x+6x.
    LAST terms (10x+3)(2x-5) for 20x^2-50x+6x-15.
    And 20x^2-50x+6x-15=20x^2-44x-15.

    Quote Originally Posted by somfooleishfool
    I find the textbook to be lacking in assumed knowledge. I found there is a whole other step needed in quadratics when a is not = 1 in ax^2. This for some reason was not explained in the textbook but I found it in my lecture notes.
    Let me thank you for not making me explain that part. You posted while I was still writing this.

    As an exercise, can you explain why setting the polynomial to zero and factoring it out into binomials provides potential solutions to the equation in the constant's of the binomial's?
    I was some of the mud that got to sit up and look around.
    Lucky me. Lucky mud.
    -Kurt Vonnegut Jr.-
    Cat's Cradle.
    Reply With Quote  
     

  21. #20  
    Forum Sophomore somfooleishfool's Avatar
    Join Date
    May 2011
    Location
    New Zealand
    Posts
    196
    Quote Originally Posted by GiantEvil
    Here goes...

    Let's start with the equation 20x^2 -44x -15 =0, and were just going to forget about the =0 part for now.

    The part were going to concern ourselves with, 20x^2 -44x -15, is referred to as a POLYNOMIAL.
    POLY means "many" or "several", "more than one". NOMIAL means "number". So polynomial means "many numbers".

    Each part of the polynomial is called a TERM. The parts or term's are [20x^2], and [-44x], and [-15].

    Each term has a COEFFICIENT which is the number part, [20x^2], [-44x], [-15]
    Each term may or may not have a VARIABLE. The variable is the letter part, [20x^2], [-44x].
    If a term has a coefficient but no variable, then that term is called a CONSTANT.
    In our particular polynomial, [-15] is a constant.

    Taking this, 20x^2-44x-15, and turning it into this, (10x+3)(2x-5), is called FACTORING.
    Factoring is when a term or polynomial is broken down into smaller parts, but only under the operation of multiplication.
    By the way, this (10x+3), and this (2x-5), are each called a BINOMIAL.
    BI means "two", and you know what "nomial" means.

    To turn our binomials, (10x+3)(2x-5), back into our polynomial, 20x^2-44x-15, we can use the F.O.I.L. method.
    FIRST we recombine the first terms(10x+3)(2x-5) for 20x^2.
    OUTSIDE terms are recombined(10x+3)(2x-5) for 20x^2-50x.
    INSIDE terms (10x+3)(2x-5) for 20x^2-50x+6x.
    LAST terms (10x+3)(2x-5) for 20x^2-50x+6x-15.
    And 20x^2-50x+6x-15=20x^2-44x-15.

    Quote Originally Posted by somfooleishfool
    I find the textbook to be lacking in assumed knowledge. I found there is a whole other step needed in quadratics when a is not = 1 in ax^2. This for some reason was not explained in the textbook but I found it in my lecture notes.
    Let me thank you for not making me explain that part. You posted while I was still writing this.

    As an exercise, can you explain why setting the polynomial to zero and factoring it out into binomials provides potential solutions to the equation in the constant's of the binomial's?
    Had to read your question and your post litterally about 5 times to understand what you were asking. So if I've interpretted you correctly (which I'm not so certain I have) then it's because you know that the variable will equal zero after you -+ the constant.
    Reply With Quote  
     

  22. #21  
    Forum Bachelors Degree x(x-y)'s Avatar
    Join Date
    Jul 2010
    Location
    UK
    Posts
    462
    Just out of curiosity, have you learnt the quadratic formula method of solving polynomial equations (well, quadratic equations) yet:

    "Nature doesn't care what we call it, she just does it anyway" - R. Feynman
    Reply With Quote  
     

  23. #22  
    Forum Bachelors Degree 15uliane's Avatar
    Join Date
    Feb 2011
    Location
    depends...
    Posts
    425
    if he did it wouldn't be a good thing-my math teacher said one has to know how to find the quadratic equation before one can learn it. We went through all the factoring, completing the square, graphing etc. before we got to use the formula.
    Reply With Quote  
     

  24. #23  
    Forum Ph.D. Heinsbergrelatz's Avatar
    Join Date
    Aug 2009
    Location
    Singapore
    Posts
    994
    Quote Originally Posted by 15uliane
    if he did it wouldn't be a good thing-my math teacher said one has to know how to find the quadratic equation before one can learn it. We went through all the factoring, completing the square, graphing etc. before we got to use the formula.
    That is correct and the same goes for any other mathematical theorem and concept. I find it surprising some people in my grade still cannot prove the quadratic formula, and yet whenever they solve quadratic equations, they go "use the the quadratic formula'. It just makes me laugh.
    ------------------




    "Mathematicians stand on each other's shoulders."- Carl Friedrich Gauss


    -------------------
    Reply With Quote  
     

  25. #24  
    Forum Bachelors Degree x(x-y)'s Avatar
    Join Date
    Jul 2010
    Location
    UK
    Posts
    462
    I know, I can prove the quadratic formula using the completing the square method- but I was just wondering whether he knew the quadratic formula yet or not (which I'm hoping not since he hasn't done completing the square).
    "Nature doesn't care what we call it, she just does it anyway" - R. Feynman
    Reply With Quote  
     

  26. #25  
    Suspended
    Join Date
    Dec 2008
    Location
    Nirgendwo und Ueberall
    Posts
    1,300
    I'm not good at math..but I wonder why Brits always refer to it as "maths"?
    Reply With Quote  
     

  27. #26  
    Veracity Vigilante inow's Avatar
    Join Date
    Oct 2009
    Location
    Austin, TX
    Posts
    3,500
    Quote Originally Posted by gottspieler
    I'm not good at math..but I wonder why Brits always refer to it as "maths"?
    It's short for mathematics, not mathematic.
    Reply With Quote  
     

  28. #27  
    Suspended
    Join Date
    Dec 2008
    Location
    Nirgendwo und Ueberall
    Posts
    1,300
    Quote Originally Posted by DrRocket
    Quote Originally Posted by somfooleishfool
    In example 20x^2 -44x-15=0

    20x^2*-15=-300x^2

    so then how do I know if its
    (20x-50)(x+6)
    or
    (20x+6)(x-50)
    20x^2 -44x-15=0 and 20x^2*-15=-300x^2 have no obvious relationship to one another and neither is (20x-50)(x+6) or (20x+6)(x-50).

    Maybe you should consider a different major.

    By inspection, 20x^2 -44x-15= (10x +3)(2x-5)
    Are you uptight, autistic, antisocial or just plain daft? You appear to believe that because you've mastered math yourself that you are entitled to insult and judge people who aren't yet as skilled as you. What makes you think that he isn't capable of majoring in Math? Perhaps after a few remedial classes he will surprise us all. I detest the attitude of defeatism that you often portray. He bravely went out on a limb here admitting his current level of mathematical ineptitude and risking ridicule. He doesn't deserve or need any. He needs encouragement, like any beginner.

    Keep at it. Never give up, "somfooleishfool".
    Reply With Quote  
     

  29. #28  
    . DrRocket's Avatar
    Join Date
    Aug 2008
    Posts
    5,486
    Quote Originally Posted by gottspieler
    Quote Originally Posted by DrRocket
    Quote Originally Posted by somfooleishfool
    In example 20x^2 -44x-15=0

    20x^2*-15=-300x^2

    so then how do I know if its
    (20x-50)(x+6)
    or
    (20x+6)(x-50)
    20x^2 -44x-15=0 and 20x^2*-15=-300x^2 have no obvious relationship to one another and neither is (20x-50)(x+6) or (20x+6)(x-50).

    Maybe you should consider a different major.

    By inspection, 20x^2 -44x-15= (10x +3)(2x-5)
    Are you uptight, autistic, antisocial or just plain daft? You appear to believe that because you've mastered math yourself that you are entitled to insult and judge people who aren't yet as skilled as you. What makes you think that he isn't capable of majoring in Math? Perhaps after a few remedial classes he will surprise us all. I detest the attitude of defeatism that you often portray. He bravely went out on a limb here admitting his current level of mathematical ineptitude and risking ridicule. He doesn't deserve or need any. He needs encouragement, like any beginner.

    Keep at it. Never give up, "somfooleishfool".
    Please go straight to hell.
    Reply With Quote  
     

  30. #29  
    . DrRocket's Avatar
    Join Date
    Aug 2008
    Posts
    5,486
    Quote Originally Posted by gottspieler
    Quote Originally Posted by DrRocket
    Quote Originally Posted by somfooleishfool
    In example 20x^2 -44x-15=0

    20x^2*-15=-300x^2

    so then how do I know if its
    (20x-50)(x+6)
    or
    (20x+6)(x-50)
    20x^2 -44x-15=0 and 20x^2*-15=-300x^2 have no obvious relationship to one another and neither is (20x-50)(x+6) or (20x+6)(x-50).

    Maybe you should consider a different major.

    By inspection, 20x^2 -44x-15= (10x +3)(2x-5)
    Are you uptight, autistic, antisocial or just plain daft? You appear to believe that because you've mastered math yourself that you are entitled to insult and judge people who aren't yet as skilled as you. What makes you think that he isn't capable of majoring in Math? Perhaps after a few remedial classes he will surprise us all. I detest the attitude of defeatism that you often portray. He bravely went out on a limb here admitting his current level of mathematical ineptitude and risking ridicule. He doesn't deserve or need any. He needs encouragement, like any beginner.

    Keep at it. Never give up, "somfooleishfool".
    Please go straight to hell.
    Reply With Quote  
     

  31. #30  
    Suspended
    Join Date
    Dec 2008
    Location
    Nirgendwo und Ueberall
    Posts
    1,300
    I live there already.
    Reply With Quote  
     

  32. #31  
    Forum Sophomore somfooleishfool's Avatar
    Join Date
    May 2011
    Location
    New Zealand
    Posts
    196
    Quote Originally Posted by x(x-y)
    Just out of curiosity, have you learnt the quadratic formula method of solving polynomial equations (well, quadratic equations) yet:

    Well the answer is yes and no.

    The teacher has gone over how to solve the square... but I was only very losely beginning to understand this before the above method was then required knowledge.

    I'm actually now quite confused as to what is what seeing as I didn't understand at the time what I was note-taking therefore in my notes several things that are completely unrelated got mingled in together.

    TBH I just have to many half understood ideas in my head now and its got me way too confused. I was just starting to think I was up to date until I went through the practise assignment (real one tomorrow on tuesday) and found that I can hardly answer a single question.

    A big issue I'm finding is that I'm missing a lot of basic algebra knowledge that wasn't covered because of its simplicity
    for example I find myself constantly refurring to the first excercise of my delta-maths textbook to find out what can and cant be done in algebraic formulas IE I didn't know (I was uncertain would be more accurate) that if both numbers either side of an equal were a negative, you can just take the negative away and alot of the transposing rules are lost to me. Its so knit pickey trying to go back and learn all those little rules, but its having a big impact on my overall success

    TBH I've given up on my assignment on tuesday. I don't have a chance in hell of passing it and its only worth 5%. I'm sick of spending hours on simple concepts trying to teach it to myself to no avail.
    Reply With Quote  
     

  33. #32  
    Forum Sophomore somfooleishfool's Avatar
    Join Date
    May 2011
    Location
    New Zealand
    Posts
    196
    where have I gone wrong here?

    solve for x and y

    5x+4y=47 .....(1)
    x=5-3y ......(2)

    I adopted the "simultaneous equations method"

    4y=47-5x .....(1) {switched 5x over the equals sign}

    y=47-5x/4 .......(1)(I don't know if I expressed this right but everything after the equals is the numerator to the denominator of 4) {divided everything by 4 to give a value for y}

    x=5-3(47-5x/4) .........(2) substituted y for value above


    4x=20-12(47-5x) {multiplied everything by 4 to get rid of the fraction}...(2)

    4x=20-564+60x {expanded brackets} ....(2)

    4x-60x=20-564 {pretty obvious} ....(2)

    -56X=-544 {simplified above equation} ....(2)

    56x=544 {removed negative symbols from both sides(this is allowed right?)} ...(2)

    x=544/56 {divide both sides by 56}

    x= 9 5/7 {and I stop the equation here because I know from looking at the answers that x ought to =11}

    What am I doing wrong? I thought I understood simultaneous equations but clearly not.

    edit: I just considered what might happen if I figured them out in reverse order and it worked. But I don't know why it ought to be different nor how you are supposed to know which number you're supposed to work with

    5x+4y=47 .....(1)
    x=5-3y ......(2)

    5x=47-4y
    x=47/5 -4y/5

    {47/5-4y/5}=5-3y
    {47-4y}=25-15y
    -4y+15y=25-47
    11y=-22
    y=-2 YAY!
    Reply With Quote  
     

  34. #33  
    Forum Sophomore somfooleishfool's Avatar
    Join Date
    May 2011
    Location
    New Zealand
    Posts
    196
    I don't expect people to continue to helping me here... but if someone feels like helping out again, I'm also stuck with

    simplify: (27x^9*y^6)^2/3

    seems simple enough but I swear the teacher hasnt even mentioned fraction exponents. I have no idea what to do with this
    Reply With Quote  
     

  35. #34  
    Forum Bachelors Degree x(x-y)'s Avatar
    Join Date
    Jul 2010
    Location
    UK
    Posts
    462
    Quote Originally Posted by somfooleishfool
    I don't expect people to continue to helping me here... but if someone feels like helping out again, I'm also stuck with

    simplify: (27x^9*y^6)^2/3

    seems simple enough but I swear the teacher hasnt even mentioned fraction exponents. I have no idea what to do with this
    What you have to remember here is that



    and that



    So, I'll just run through the simplification and hopefully you'll understand it:







    So, that is the final simplified expression.
    "Nature doesn't care what we call it, she just does it anyway" - R. Feynman
    Reply With Quote  
     

  36. #35  
    Suspended
    Join Date
    Dec 2008
    Location
    Nirgendwo und Ueberall
    Posts
    1,300
    You need to take some remedial courses, then you'll be fine.
    Reply With Quote  
     

  37. #36  
    Forum Ph.D. Heinsbergrelatz's Avatar
    Join Date
    Aug 2009
    Location
    Singapore
    Posts
    994
    somefoolishfool, what are you majoring in??
    ------------------




    "Mathematicians stand on each other's shoulders."- Carl Friedrich Gauss


    -------------------
    Reply With Quote  
     

  38. #37  
    . DrRocket's Avatar
    Join Date
    Aug 2008
    Posts
    5,486
    Quote Originally Posted by gottspieler
    You need to take some remedial courses, then you'll be fine.
    Thisv stuff IS remedial, and very low level remedial at that.
    Reply With Quote  
     

Bookmarks
Bookmarks
Posting Permissions
  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •