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secondary 3 | A Maths
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Date Posted: 4 years ago
Views: 634
Hi! the equal sign is actually 3 lines not 2 and the working of x is mine so it's not part of the question tysm!
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Sorry! I dont really understand the workings written
Can you be more specific which part of the working is not clear?
The 3 line equal sign means "is identical to".
For example, when we write 2x = 8, this is true only when x = 4.
However, when we write 2x = 6x/3, we use 3 line equal sign because the statement is true for all values of x.
For this question, the 3 line equal sign is used because the statement is true for all values of x. We should follow and use the 3 line equal sign. Apologies my oversight.
The only maybe confusing part I can see is how to know that Q(x) must be in the form of 2x+c.
On LHS we have 2x^3 ….
On RHS we have (x - 1)(x+2) multiply by "something".
The only way that LHS can be identical to RHS is that the "something" must contain a "2x" inside, so that you get …
"x" times "x" times "2x" equals "2x^3".
Because LHS is identical to RHS, so the coefficients of x^3, x^2, x and the constant term on the 2 sides must be equal. By equating the coefficients, we can solve for a, b and c.
The 3 line equal sign means "is identical to".
For example, when we write 2x = 8, this is true only when x = 4.
However, when we write 2x = 6x/3, we use 3 line equal sign because the statement is true for all values of x.
For this question, the 3 line equal sign is used because the statement is true for all values of x. We should follow and use the 3 line equal sign. Apologies my oversight.
The only maybe confusing part I can see is how to know that Q(x) must be in the form of 2x+c.
On LHS we have 2x^3 ….
On RHS we have (x - 1)(x+2) multiply by "something".
The only way that LHS can be identical to RHS is that the "something" must contain a "2x" inside, so that you get …
"x" times "x" times "2x" equals "2x^3".
Because LHS is identical to RHS, so the coefficients of x^3, x^2, x and the constant term on the 2 sides must be equal. By equating the coefficients, we can solve for a, b and c.
done
{{ upvoteCount }} Upvotes
clear
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required a substitution to make it clearer
Date Posted:
4 years ago
mistake in the x^2 term.
instead of … + (1+d)x^2+ … ,
should be … + (c+d)x^2+...
so, comparing x^2 term,
-4 = c+d
since c=2
therefore d=-6 (not -5)
continuing on, a = 5, b=-14.
instead of … + (1+d)x^2+ … ,
should be … + (c+d)x^2+...
so, comparing x^2 term,
-4 = c+d
since c=2
therefore d=-6 (not -5)
continuing on, a = 5, b=-14.
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