Sec3 AMaths
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Factorise - 2 - 7x - 2x² + 3x³ completely. Hence, solve 3x - 4x² = 28x³ + 16x^4 .
I don't know how to do this. Can someone help me? Thanks
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Hi,
Notice that x + 1 is a factor of - 2 - 7x - 2x² + 3x³.
The second expression does not seem to resemble the first in any way. Usually it should, because of 'Hence'. Do check it out.
Thanks!
Cheers,
Wen Shih -
Originally posted by wee_ws:
Hi,
Notice that x + 1 is a factor of - 2 - 7x - 2x² + 3x³.
The second expression does not seem to resemble the first in any way. Usually it should, because of 'Hence'. Do check it out.
Thanks!
Cheers,
Wen Shih
My roots for the 1st equation are (x + 1), (x - 2) & (3x + 1). I believe that is the easiest part so not much explanation is needed. And I agree with the mod on the 2nd expression not resembling the 1st in any way. They don't even have the same roots! Maybe you copied the question wrongly??? -
Originally posted by Only-Way-4-Destiny!:
Factorise - 2 - 7x - 2x² + 3x³ completely. Hence, solve 3x - 4x² = 28x³ + 16x^4 .
I don't know how to do this. Can someone help me? Thanks
- 2 - 7x - 2x² + 3x³
=(x-2)(x+1)(1+3x)
3x - 4x² = 28x³ + 16x^4
3x - 4x² -28x³ -16x^4=0
x^4(3x^-3-4x^-2 - 28x^-1 -16)=0
x^4 = 0 or 3x^-3-4x^-2 - 28x^-1 -16=0
let x^-1 = 2y
3(8y^3)-4(4y^2)-28(2y)-16=0 .... divide by 8
3y^3-2y^2-7y-2=0
(y-2)(y+1)(3y+1)=0
y = 2, -1 or -1/3
2y = 4, -2 or -2/3
x^-1 = 4, -2 or -2/3
x = 1/4, -1/2 or -3/2
Therefore x = 0, 1/4, -1/2 or -3/2 -
Wow, interesting. Didn't know nowadays sec 3 A maths requires so much work...
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Hi,
An eye-opener, although it is too much to ask for from a sec 3 student.
Thanks!
Cheers,
Wen Shih -
My teacher said the question was correct, the question is correct and it was a killer question. So idk how to do. Perhaps need substituion?
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Originally posted by Only-Way-4-Destiny!:
My teacher said the question was correct, the question is correct and it was a killer question. So idk how to do. Perhaps need substituion?
Hi I have provided the solution already. Is it unclear in any way? Perhaps you can tell me which part of the solution you do not understand and I can explain it.
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Originally posted by Only-Way-4-Destiny!:
My teacher said the question was correct, the question is correct and it was a killer question. So idk how to do. Perhaps need substituion?
Scroll up. Mikethm already posted the correct answer.
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I'm more curious on how many marks this question is worth.
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Not stated on my paper, the marks
Anyway, I have manipulated the hence part.
3x - 4x² = 28x³ + 16x^4
3x = 16x^4 + 28x³ + 4x²
3x = 4x(4x³ + 7x² + x)
3x/4x = 4x³ + 7x² + x
3/4 = 4x³ + 7x² + x
16x³ + 28x² + 4x - 3 = 0
I got the 2nd cubic equation but I am supposed to use the first part to solve. I solved the equation without the first part. I got
x = 0.25, -1.5 or -0.5
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its x^4 so you need 4 values of x. U missed out x = 0
The correct method is posted already anyway
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This is the alternative method of solving but without using the first part. There's "HENCE" , making the question damn hard
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Originally posted by Mikethm:
- 2 - 7x - 2x² + 3x³
=(x-2)(x+1)(1+3x)
3x - 4x² = 28x³ + 16x^4
3x - 4x² -28x³ -16x^4=0
x^4(3x^-3-4x^-2 - 28x^-1 -16)=0
x^4 = 0 or 3x^-3-4x^-2 - 28x^-1 -16=0
let x^-1 = 2y
3(8y^3)-4(4y^2)-28(2y)-16=0 .... divide by 8
3y^3-2y^2-7y-2=0
(y-2)(y+1)(3y+1)=0
y = 2, -1 or -1/3
2y = 4, -2 or -2/3
x^-1 = 4, -2 or -2/3
x = 1/4, -1/2 or -3/2
Therefore x = 0, 1/4, -1/2 or -3/2
Wow! That's a really good job in spotting the pattern. =D -
Originally posted by Only-Way-4-Destiny!:
This is the alternative method of solving but without using the first part. There's "HENCE" , making the question damn hard
x^4(3x^-3-4x^-2 - 28x^-1 -16)=0
x^4 = 0 or 3x^-3-4x^-2 - 28x^-1 -16=0
let x^-1 = 2y
3(8y^3)-4(4y^2)-28(2y)-16=0 .... divide by 8
3y^3-2y^2-7y-2=0
This is the part where you use the hence. Instead of slowly working out the factors, you just throw (y-2)(y+1)(3y+1)=0 back to them directly. This is what is used from the 1st question.
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Originally posted by Mikethm:
- 2 - 7x - 2x² + 3x³
=(x-2)(x+1)(1+3x)
3x - 4x² = 28x³ + 16x^4
3x - 4x² -28x³ -16x^4=0
x^4(3x^-3-4x^-2 - 28x^-1 -16)=0
x^4 = 0 or 3x^-3-4x^-2 - 28x^-1 -16=0
let x^-1 = 2y
3(8y^3)-4(4y^2)-28(2y)-16=0 .... divide by 8
3y^3-2y^2-7y-2=0
(y-2)(y+1)(3y+1)=0
y = 2, -1 or -1/3
2y = 4, -2 or -2/3
x^-1 = 4, -2 or -2/3
x = 1/4, -1/2 or -3/2
Therefore x = 0, 1/4, -1/2 or -3/2Wow! The answer was here! Very impressive. Thanks a lot, Mikethm!
Impressive
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Glad to be of service.
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your alternative method miss out x=0 because you take 3x / 4x = 3/4. This is BOOBOO!
Mikethm, how did you spot the pattern? perhaps can share
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maybe he set the question?
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@SBS261P
Ignoring the x 1st and concentrating on how to get the coefficients to match. Once that is done, you will see that the "x"s are in the opposite order of what they should be. Then it is a matter of flexibility.
@HiAy3Captain
I am offended that you think I was the one who set this stupid question which require a unnecessarily complicated solution to a simple question. :)
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Originally posted by SBS261P:
your alternative method miss out x=0 because you take 3x / 4x = 3/4. This is BOOBOO!
Mikethm, how did you spot the pattern? perhaps can share
Good job by Mikethm! I can't think of a simpler way than his.
One way to spot the pattern could be to take note:
3x³ - 2x² - 7x - 2
and
3x^-3 - 4x^-2 - 28x^-1 -16
Note 3*2^0 = 3, 2*2^1 = 4, 7*2^2=28. 2*2^3 = 16
From there, you know the pattern will roughly require the number 2 somewhere in the substitution.A slight variation/alternative method:
16x^4 + 28x³ + 4x² - 3x = 0
x(16x³ + 28x² + 4x - 3) = 0Either x = 0 or 16x³ + 28x² + 4x - 3 = 0
for 16x³ + 28x² + 4x - 3, sub x = 1/(2y) ==> it's the same as Mikethm's of subbing x^-1 = 2y
2y^-3 + 7y^-2 + 2y^-1 - 3 = 0
Multiply by -y³ throughout.
-2 - 7y - 2y² + 3y³ = 0
From here on same as Mikethm's method.