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    (Original post by SeanFM)
    Right.. and how have you solved them / why do you think its taking so much time? It could be that you're just getting used to it at the start.
    It's not taking too long, I just want to see if there is a faster way lol.
    I'm quite fast at calculating, so it's not too long tbh.

    EDIT: Latex is being a pain in the ass, look at (g) I've used the same thing for (h) I'll put up in around 3 mins.
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    (Original post by SeanFM)
    Right.. and how have you solved them / why do you think its taking so much time? It could be that you're just getting used to it at the start.
    Put up both methods.
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    (Original post by Chittesh14)
    Put up both methods.
    It's on the previous page... :rolleyes: :hide: :rofl: okay.
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    (Original post by Chittesh14)
    Questions: Solve, giving your answer to 3 s.f.

    (g) 32x+1 - 26(3x) - 9 = 0
    (h) 4(32x+1) + 17(3x) - 7 = 0

    How I solved them:

    (g)
    3^{2x+1} - 26(3^x) - 9 = 0

3^{2x} - 26(3^{x-1}) - 3 = 0

3^{2x} - 26 \frac{3^x}{3^1} - 3 = 0

Let y = 3^x

y^2 - \frac{26y}{3} - 3 = 0

3y^2 - 26y - 9 = 0

3y^2 + y - 27y - 9 = 0

3y(y+1/3) - 27(y+1/3) = 0

(3y-27)(y+1/3) = 0

(y-9)(3y+1) = 0

y = 9 or y = -1/3



When y = 9, 3^x = 9
    log39 = x
    x = 2

    When y = -1/3, there are no solutions.

    (h)
    4(3^{2x+1}) - 17(3^x) - 7 = 0

4(3^{2x}) + 17(3^{x-1}) - 7/3 = 0

Let y = 3^x

4y^2 - \frac{17y}{3} - 7/3 = 0

12y^2 - 17y - 7 = 0

12y^2 - 4y + 21y - 7 = 0

12y(y-1/3) + 21(y-1/3) = 0

(12y+21)(y-1/3) = 0

(4y+7)(3y-1) = 0

y = -7/4 or y = 1/3



When y = 1/3, 3^x = 1/3
    log31/3 = x
    x = -1

    When y = -7/4, there are no solutions.
    I see why you feel like it's longer than it should be - you've divided by 3 and then multiplied by 3 again.

    When you bring the 3 down in 3^(2x+1), leave it in front, so you get 3*(3^2x) and then substitute as normal. Otherwise, nothing else you can do to speed it up.
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    (Original post by SeanFM)
    I see why you feel like it's longer than it should be - you've divided by 3 and then multiplied by 3 again.

    When you bring the 3 down in 3^(2x+1), leave it in front, so you get 3*(3^2x) and then substitute as normal. Otherwise, nothing else you can do to speed it up.
    Oh right, amazing. I don't even think it was taught in the book like this, I just use my own methods lmao if I feel they work. Thanks a lot! .
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    (g)

     3^{2x+1} - 26(3^x) - 9 = 0

3{(3^x)^2}- 26(3^x) - 9 = 0
    Treat this as a quadratic expression, you can let u=3x if you want wish to in order for it to appear simpler. Then factorise it.

     (3^x- 9)(3(3^x) + 1) = 0

    Therefore  3^x = 9 or -1/3 ; disregard -1/3 as  3^x>0 for all x
    -> 3^x=9 therefore x=2 from inspection (I guess you can use logs just to show it's true)

    (h)

    4(3^{2x+1}) + 17(3^x) - 7 = 0

12{(3^x)^2} + 17(3^x) - 7 = 0

    Treat it as a quadratic again and factorise it.

     (3(3^x) - 1)(4(3^x) + 7) = 0

    Therefore  3^x=1/3 or -7/4 ; disregard negative
    ->  3^x=1/3 therefore x=-1 from inspection

    The only key thing here is to realise a^{(b+c)}={a^b}{a^c} and a^{(2b)}=(a^b)^2
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    (Original post by RDKGames)
    (g)

     3^{2x+1} - 26(3^x) - 9 = 0

3{(3^x)^2}- 26(3^x) - 9 = 0
    Treat this as a quadratic expression, you can let u=3x if you want wish to in order for it to appear simpler. Then factorise it.

     (3^x- 9)(3(3^x) + 1) = 0

    Therefore  3^x = 9 or -1/3 ; disregard -1/3 as  3^x>0 for all x
    -> 3^x=9 therefore x=2 from inspection (I guess you can use logs just to show it's true)

    (h)

    4(3^{2x+1}) + 17(3^x) - 7 = 0

12{(3^x)^2} + 17(3^x) - 7 = 0

    Treat it as a quadratic again and factorise it.

     (3(3^x) - 1)(4(3^x) + 7) = 0

    Therefore  3^x=1/3 or -7/4 ; disregard negative
    ->  3^x=1/3 therefore x=-1 from inspection

    The only key thing here is to realise a^{(b+c)}={a^b}{a^c} and a^{(2b)}=(a^b)^2
    Wow, this is what I was looking for.
    Thank you so much ♥, I'm going to do logs again later on today after I stop playing games!!
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    (Original post by RDKGames)
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    (Original post by SeanFM)
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    Does this statement: "You can divide a polynomial by (x \pm p)." mean that you can divide a polynomial by (x (a positive or negative number) so e.g. (x-2) or (x+2) right? I feel I misinterpreted it when I thought it meant you can divide any polynomial by p as a negative or positive number so again (x-2) and (x+2).

    I know that you probably won't get what I'm saying lool, but I hope you do.
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    (Original post by Chittesh14)
    Does this statement: "You can divide a polynomial by (x \pm p)." mean that you can divide a polynomial by (x (a positive or negative number) so e.g. (x-2) or (x+2) right? I feel I misinterpreted it when I thought it meant you can divide any polynomial by p as a negative or positive number so again (x-2) and (x+2).

    I know that you probably won't get what I'm saying lool, but I hope you do.
    I'm not entirely sure what the difference is between what you've said and what they've said.

    If you fix a p value, whether it's positive or negative, the + / - will cover both ends.

    But you're getting far too bogged down in the specification and such statements :lol: what's important is that you can divide by (x+a) or (x-a) where a is a positive constant.
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    (Original post by SeanFM)
    I'm not entirely sure what the difference is between what you've said and what they've said.

    If you fix a p value, whether it's positive or negative, the + / - will cover both ends.

    But you're getting far too bogged down in the specification and such statements :lol: what's important is that you can divide by (x+a) or (x-a) where a is a positive constant.
    Lol, no what I was thinking was:

    If a polynomial is let's say... x^3 - 10x^2 + 19x + 30, since (x+1) is a factor, I was thinking that if p = 1, (x-1) is also a factor. But, that is incorrect because if you factorise it, you get: (x+1)(x-5)(x-6) lol.
    I hope you understand what I am saying now .
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    (Original post by Chittesh14)
    Lol, no what I was thinking was:

    If a polynomial is let's say... x^3 - 10x^2 + 19x + 30, since (x+1) is a factor, I was thinking that if p = 1, (x-1) is also a factor. But, that is incorrect because if you factorise it, you get: (x+1)(x-5)(x-6) lol.
    I hope you understand what I am saying now .
    Ohh, I see what you mean. but yeah - just go with what I said about 'a' in my previous post.
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    (Original post by Chittesh14)
    Does this statement: "You can divide a polynomial by (x \pm p)." mean that you can divide a polynomial by (x (a positive or negative number) so e.g. (x-2) or (x+2) right? I feel I misinterpreted it when I thought it meant you can divide any polynomial by p as a negative or positive number so again (x-2) and (x+2).I know that you probably won't get what I'm saying lool, but I hope you do.
    (Original post by Chittesh14)
    Lol, no what I was thinking was:If a polynomial is let's say... x^3 - 10x^2 + 19x + 30, since (x+1) is a factor, I was thinking that if p = 1, (x-1) is also a factor. But, that is incorrect because if you factorise it, you get: (x+1)(x-5)(x-6) lol.I hope you understand what I am saying now .
    I think I know what you mean and I see the flaw in your thought. Let's say you have some polynomial p(x).

    The book is basically saying you CAN DIVIDE the polynomial by any linear expression such as (x \pm a) however this does NOT mean x=a and x=-a are BOTH roots of p(x)=0, therefore it does NOT mean (x-a) and (x+a) are both factors of p(x). ONLY if \frac{p(x)}{(x-a)} gives a remainder of 0 will x=a be a solution to it equaling zero. (or x=-a if divided through by (x+a))

    Just because you can divide a polynomial by a linear expression, it does not strictly mean that linear expression has a root for the polynomial.

    Practice long division with polynomial expressions to see what I mean.
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    Just realised p(n) should be written as p(x) instead. Wouldn't make sense for a function of n to be expressed in terms of x, sorry. xD
    But this is what I mean, therefore x=2 a real root if p(x)=0. If you plug in x=-2, the first division will not be determined because you would be dividing by 0 on the denominator, thus x=-2 cannot be a root.

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    In all honesty, I knew where I was wrong and what it meant. I just wanted to see you how both would react to it and explain it to me !

    (Original post by SeanFM)
    Ohh, I see what you mean. but yeah - just go with what I said about 'a' in my previous post.
    Yeah , I realised that haha. Thanks for explaining it .

    (Original post by RDKGames)
    I think I know what you mean and I see the flaw in your thought. Let's say you have some polynomial p(x).

    The book is basically saying you CAN DIVIDE the polynomial by any linear expression such as (x \pm a) however this does NOT mean x=a and x=-a are BOTH roots of p(x)=0, therefore it does NOT mean (x-a) and (x+a) are both factors of p(x). ONLY if \frac{p(x)}{(x-a)} gives a remainder of 0 will x=a be a factor of it. (or x=-a if divided through by (x+a))

    Just because you can divide a polynomial by a linear expression, it does not strictly mean that linear expression has a root for the polynomial.

    Practice long division with polynomial expressions to see what I mean.
    (Original post by RDKGames)
    Just realised p(n) should be written as p(x) instead. Wouldn't make sense for a function of p to be expressed in terms of x, sorry. xD
    But this is what I mean, therefore x=2 a real root if p(x)=0. If you plug in x=-2, the first division will not be determined because you would be dividing by 0 on the denominator, thus x=-2 cannot be a root.

    Name:  ImageUploadedByStudent Room1466881529.386010.jpg
Views: 56
Size:  141.7 KB

    Don't worry, I've done a lot of long divsion :P. Loads of questions and practice, mastered it. I just got a bit confused on a different question and realised this, hence why I asked. Thanks for the explanation though .
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    (Original post by Chittesh14)
    In all honesty, I knew where I was wrong and what it meant. I just wanted to see you how both would react to it and explain it to me !



    Yeah , I realised that haha. Thanks for explaining it .
    Well lucky you I was bored enough to explain it detail, otherwise I would've left SeanFM's explanation for you :P
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    (Original post by RDKGames)
    Well lucky you I was bored enough to explain it detail, otherwise I would've left SeanFM's explanation for you :P
    Lol . I just like some explanations lol, just to understand it perfectly. Anyway, long divsion is easy, got to move on tomorrow! For today - it's just TV - football and boxing .
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    (Original post by Chittesh14)
    In all honesty, I knew where I was wrong and what it meant. I just wanted to see you how both would react to it and explain it to me !
    Riiight. Thanks for that ... :toofunny:
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    OP please do me a favour and solve x2+2x+2=0 , I'm so stuck on it!
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    (Original post by RDKGames)
    OP please do me a favour and solve x2+2x+2=0 , I'm so stuck on it!
    Erm, well I can't really factorise it so let's just go with my favourite method lol.
    (x+1)^2 - 1 + 2 = 0

(x+1)^2 + 1 = 0

(x+1)^2 = -1

    Well, you can't complete the square either .

    Well, the quadratic formula doesn't work either wtf lol, this equation has no solutions.
    b^2 - 4ac = 4 - 8 = -4

b^2 - 4ac < 0 so there are no real roots.

    Apparently, the solutions (roots) are imaginary lol so the graph does not cross the x-axis.
    I tried WolframAlpha and it says -1 + i and -1 - i.
    I don't know what that means lol :/ sorry I can't help :'(.

    Btw, what year are you in? - just wondering
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    (Original post by SeanFM)
    Riiight. Thanks for that ... :toofunny:
    Haha, sorry for wasting your time . But, at least it helped me - so it wasn't really wasted .
 
 
 
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