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Student 999
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Struggling on the last part, it seems to me I need to translate it somehow to remove the y^3 coefficient but can't see how![Name: Screenshot 2022-07-02 at 16.20.35.png
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Last edited by Student 999; 1 month ago
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mqb2766
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#2
(Original post by Student 999)
Struggling on the last part, it seems to me I need to translate it somehow to remove the y^3 coefficient but can't see how![Name: Screenshot 2022-07-02 at 16.20.35.png
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Struggling on the last part, it seems to me I need to translate it somehow to remove the y^3 coefficient but can't see how
Think of a translation like
z-a = y
so that when you sub it in, the cubic term disappears. Its obviously related to the cubic term in a 4th power binomial.
Last edited by mqb2766; 1 month ago
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Student 999
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(Original post by mqb2766)
Its the "same" as when you complete the square of a quadratic, or remove the quadratic term from a cubic.
Think of a translation like
z-a = y
so that when you sub it in, the cubic term disappears. Its obviously related to a 4th power binomial.
Its the "same" as when you complete the square of a quadratic, or remove the quadratic term from a cubic.
Think of a translation like
z-a = y
so that when you sub it in, the cubic term disappears. Its obviously related to a 4th power binomial.
Is the reason why I can do that because for a general quartic x^4 +bx^3 + cx^2 + dx +e can be rewritten as (x-a_1)^4 + f(x-a_2)^2 + a_3 so a translation will affect the x^3 term of the first bracket (x-a_1)^4 whereas the rest of the brackets do not contain a x^3 term.
If I wanted to remove the x^2 term though how would I go about that, from the identity that I came up with I would need to somehow adjust the a_1 constant to produce a quartic with x^2 coefficient of -f to cancel out the quadratic term in f(x-a_2)^2 ?
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(Original post by Student 999)
Is the reason why I can do that because for a general quartic x^4 +bx^3 + cx^2 + dx +e can be rewritten as (x-a_1)^4 + f(x-a_2)^2 + a_3 so a translation will affect the x^3 term of the first bracket (x-a_1)^4 whereas the rest of the brackets do not contain a x^3 term.
If I wanted to remove the x^2 term though how would I go about that, from the identity that I came up with I would need to somehow adjust the a_1 constant to produce a quartic with x^2 coefficient of -f to cancel out the quadratic term in f(x-a_2)^2 ?
Is the reason why I can do that because for a general quartic x^4 +bx^3 + cx^2 + dx +e can be rewritten as (x-a_1)^4 + f(x-a_2)^2 + a_3 so a translation will affect the x^3 term of the first bracket (x-a_1)^4 whereas the rest of the brackets do not contain a x^3 term.
If I wanted to remove the x^2 term though how would I go about that, from the identity that I came up with I would need to somehow adjust the a_1 constant to produce a quartic with x^2 coefficient of -f to cancel out the quadratic term in f(x-a_2)^2 ?
If you had a quartic and wanted to remove the quadratic term (so keeping the quartic, cubic, linear and constant terms) you could do something similar, just expand the transformed quartic and cubic binomials and pick the "a" which makes the quadratic term zero. I don't think this is part of the question? Note you couldn't do this on both the cubic and quadratic terms. You have one knob to turn "a", so you'll only be able to control one term being zero. Note, not worked through the corresponding equation though.
Last edited by mqb2766; 1 month ago
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