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Nady [450]
2 years ago
7

how do I solve this quadratic simultaneous equation 2y+2x=7 and x^2-4y^2=8 it has two sets of solutions and I'm solving by both

variables

Mathematics
2 answers:
Maksim231197 [3]2 years ago
8 0
Sent a picture of the solution to the problem (s).

Ghella [55]2 years ago
3 0
These are all the possible answers i know from experience.

2x + 2y = 7
x^2 - 4y^2 = 8

2x = -2y + 7
x = -y + 7/2

(-y + 7/2)^2 - 4y^2 = 8

y^2 - 7y + 49/4 - 4y^2 = 8

0 = 3y^2 + 7y + 8 - 49/4

0 = 12y^2 + 28y + 32 - 49

0 = 12y^2 + 28y - 17

y = (-b ± √(b^2 - 4ac))/(2a)

y = (-28 ± √(28^2 - 4(12)(-17)))/(2(12))

Dividing by 2 is like dividing by √(4):

y = (-14 ± √(7*28 + (12)(17)))/(2*6)

y = (-7 ± √(7*7 + (3)(17)))/6

y = (-7 ± 10)/6 = 1/2 or -17/6

x = -y + 7/2 = 3 or 19/3

<span>Solutions are (3,1/2) and (19/3,-17/6)</span>
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Answer:

Step-by-step explanation:


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Now in this case you'll notice the degree is still even (it's 4) and the 4 is also a perfect square, and it's a difference of squares in one of the factors, so it can further be rewritten:

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I'm assuming that's considered completely factored but you can technically factor it further. While the identity difference of squares technically only applies to difference of squares, it can also be used on the sum of squares, but you need to use imaginary numbers. Because x^2+4 = x^2-(-4). and in this case a=x^2 and b=-4. So rewriting it as the difference of squares becomes: x^4+4 = x^4 - (-4) = (x^2-\sqrt{-4})(x^2+\sqrt{-4}) = (x^2-2i)(x^2+2i) just something that might be useful in some cases.

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