Help with Algebra II/Trig problem!
- Thread starter farhanaq
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I just put the periods there to seperate the problems so you can see it better
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Is the above meant to be as follows?
. . . . .\(\displaystyle \dfrac{-8y\, +\, 3x}{4x^2\, -\, 2xy^2\, +\, y^4}\, +\, \dfrac{x}{2xy^2\, -\, 4x^2\, -\, y^4}\)
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When you reply, please include the instructions for the expression, along with a clear listing of what you've tried so far. Thank you!
HallsofIvy
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In order to add fractions, you need to get a "common denominator". And in order to do that, you have to see what factors the denominator might already have in common. So start by factoring the denominators.
\(\displaystyle 4x^2- 2xy^2+ y^4= 2(x^2- xy^2)+ y^4= 2(x^2- xy^2+ (1/4)y^4- (1/4)y^4)+ y^4\)
(I completed the square in the first two terms)
\(\displaystyle = 2(x^2- xy^2+ (1/4)y^4)- (1/2)y^4+ y^4= 2(x- (1/2)y^2)+ (1/2)y^4\)
Since that is a sum of squares, it cannot be factored. It then does not matter whether the other denominator can be factored. Since the first one cannot be factored, they cannot have any common factors.
So all we can do to get a common denominator is to multiply both numerator and denominator of each fraction by the denominator of the other.
\(\displaystyle \frac{(8x+ 3y)(2x^2y- 4x^2+ y^4)+ x(4x^2- 2xy+ y^4)}{(2x^2y- 4x^2+ y^4)(4x^2- 2xy+ y^4)}\)
Can you finish that?
\(\displaystyle 4x^2- 2xy^2+ y^4= 2(x^2- xy^2)+ y^4= 2(x^2- xy^2+ (1/4)y^4- (1/4)y^4)+ y^4\)
(I completed the square in the first two terms)
\(\displaystyle = 2(x^2- xy^2+ (1/4)y^4)- (1/2)y^4+ y^4= 2(x- (1/2)y^2)+ (1/2)y^4\)
Since that is a sum of squares, it cannot be factored. It then does not matter whether the other denominator can be factored. Since the first one cannot be factored, they cannot have any common factors.
So all we can do to get a common denominator is to multiply both numerator and denominator of each fraction by the denominator of the other.
\(\displaystyle \frac{(8x+ 3y)(2x^2y- 4x^2+ y^4)+ x(4x^2- 2xy+ y^4)}{(2x^2y- 4x^2+ y^4)(4x^2- 2xy+ y^4)}\)
Can you finish that?
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There is no need for that. The denominators are opposites in sign of each other. \(\displaystyle \ \ \ \ \)\(\displaystyle \dfrac{-8y + 3x}{4x^2 - 2xy^2 + y^4} \ + \ \dfrac{x}{2xy^2 - 4x^2 - y^4} \ = \ \)
\(\displaystyle \ \ \ \ \)\(\displaystyle \dfrac{-8y + 3x}{4x^2 - 2xy^2 + y^4} \ + \ \dfrac{x}{- 4x^2 + 2xy^2 - y^4} \ = \ \)
\(\displaystyle \ \ \ \ \)\(\displaystyle \dfrac{-8y + 3x}{4x^2 - 2xy^2 + y^4} \ + \ \dfrac{(-1)x}{(-1)(- 4x^2 + 2xy^2 - y^4)} \ = \ \)
\(\displaystyle \ \ \ \ \)\(\displaystyle \dfrac{-8y + 3x}{4x^2 - 2xy^2 + y^4} \ + \ \dfrac{-x}{4x^2 - 2xy^2 + y^4} \ = \ \)
\(\displaystyle \ \ \ \ \ \ \ \ \)
Now continue from that.
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