# simultaneous equations, specially for youWatch

This discussion is closed.
#1
x +√ y = a
√x + y = b
0
#2
well i gave up and got mathematica to solve

0
14 years ago
#3
x=2 tbh
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14 years ago
#4
what the **** was all that?
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14 years ago
#5
lmao that's great
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14 years ago
#6
(Original post by fishpaste)
x +√ y = a
√x + y = b

Do you want x and y in terms of a and b?
0
14 years ago
#7
ok im really worried now- is this degree or a level maths? and if its a level maths then is it p6?
0
14 years ago
#8
{x + √y = a} => {x = a - √y}
√x + y = b

√(a - √y) + y = b

(a - √y), consider this is a perfect square. that is,
(a - √y) = (c + d)Â² = cÂ² + 2cd + dÂ²

if we put:

{-√y = 2cd} => {y = 4cÂ²dÂ²}
a = cÂ² + dÂ²

Now a is the sum of cÂ² and dÂ², and y is the product of 4, cÂ² and dÂ²
We can use a quadratic to show this:

zÂ² + az + (y/4) = (z + cÂ²)(z + dÂ²) = 0

Erm... someone care to take over, I got distracted and forgot where I was heading.
0
14 years ago
#9
(Original post by fishpaste)
x +√ y = a
√x + y = b
The solution to that makes me wish I were actually dead... There is clearly no point in doing maths, if you have to do things like that... tisk.
0
14 years ago
#10
I think this was supposed to be a joke...?
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#11
um it wasn't meant to be a joke because i genuinely thought there might be a nice solution, then i saw mathematica's solution and had to post for sheer hilarity

also, there might be a simple one, mathematica can make simple things ridiculous in my experience

yep ralf, x and y in terms of a & b
0
14 years ago
#12
x+√y=a (1)
√x+y=b (2)
Thus from (1): x=a-√y
In (2): √(a-√y)+y=b
=> √(a-√y)=b-y
=>a-√y=(b-y)^2
=>√y=(a-(b-y))^2
=>y=a^2-2a(b-y)^2+(b-y)^4

Multiplying out and taking the y across:

y^4-4by^3+(6b^2-2a)y^2+(4ab-4b^3-1)y+b^4+a^2-2ab=0

However, I lack the time and the greatness to solve that right now, but if you do, then just substitute this in for the value of y in equation 1 and voila (you also have to knock out the spurious answers you've created by squaring though, so be careful...)
Or alternatively look here:
l l
l l
l l
\ /
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14 years ago
#13
A slightly nicer quartic than meepmeeps can be gotten by writing x = p^2, y = q^2 (valid since x,y >= 0)

p^2 + q = a
q^2 + p = b

Hence (b-q^2)^2 + q - a = 0 => q^4 - 2bq^2 + q + b^2 - a = 0.
0
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