Question

How do I solve for the variables here?

a^2+bc=0
ab+bd=-1
ac+cd=1
bc+d^2=0

Answer 1

Actually, I'd like a more general solution if possible where the variables a, b, c, and d are in terms of w, x, y, and z like this if possible:

a^2+bc=w
ab+bd=x
ac+cd=y
bc+d^2=z

Answer 2

the key is to consider (I think): $$(a+b+c+d)^2=a^2+b^2+c^2+d^2+2(ab+ac+ad+bc+bd+cd)$$

Answer 3

\(ab+bd=-1,ac+cd=1,a^2+bc=0,bc+d^2=0\) hence:$$(a+b+c+d)^2=b^2+c^2+2ad$$not sure what to do next... :-p

Answer 4

Hmm that might be one route to the solution. See, I already know the answer, and the way I got there just doesn't seem like the easiest. Here's what I did:

Just a little quick plugging in from the equations
a^2=d^2
b(a+d)=-1
c(a+d)=1
b=-c

since a^2+bc=0 and we know b=-c, then: a^2=b^2=c^2=d^2.

But we don't know what the signs on any of them are because they're all squared. However back above we have:

c(a+d)=1. Since we know the magnitudes of all of them are the same, a+d has to mean that a and d have the same sign, otherwise that statement would be c(0)=1 which is false. So,

a=d and still b=-c

now what about the relation between the signs of a and d with the signs of b and c? We don't know if a, b, and d are all negative/positive or if it's a,c, and d that are all negative/positive. But actually we do if we look here:

c(a+d)=1

Since a and d have the same sign, that term is either positive or negative. c must have the same sign as a and d since (-1)(-1)=(1)(1).

SO we now know:

a=-b=c=d

c(a+d)=1 now yields:

a(a+a)=1

2a^2=1

a=sqrt(2)/2

and that's what all of them are except b, which is of course negative sqrt2/2.

But this just doesn't seem like the easiest path.

Answer 5

well once you conclude \(a^2=d^2,b=-c\) we have:$$(a+b+c+d)^2=b^2+c^2+2ad\\(a+d)^2=2b^2+2ad\\a^2+d^2+2ad=2b^2+2ad\\a^2+d^2=2b^2\\2a^2=2b^2$$ hence \(a=b\text{ or }a=-b\) and \(a=d\text{ or }a=-d\)... my guess is to consider these possibilities to get the solutions

Answer 6

$$ab+bd=-1\\
ac+cd=1$$assuming \(a=b=-c=d\):$$a^2+a^2=-1$$so this is not possible.

Answer 7

so we know either \(a=-b\) or \(a=-d\)... consider \(a=-b=c=-d\):$$-a^2+a^2=-1$$so this is also not possible. consider \(a=-b=c=d\):$$-a^2-a^2=-1\\-2a^2=-1\\a^2=1/2$$hence \(a=\pm1/\sqrt2\) and this yields a couple solutions. you can try with the other combinations but they will similarly fail those two equations above

Answer 8

Lol ok, so you're pretty much saying what I just got done saying. Is there no other way?

Answer 9

Hey don't be rude, he took 30 minutes to write all that !

Answer 10

by the way, nice @oldrin.bataku I recognize a brilliant.org techniques when I see one

Answer 11

@kainui using the factored-form equations \(b(a+d)=-1,c(a+d)=1\) it's pretty trivial to determine \(a=d\) which means we're done rather quickly (since \(a=-d\) would fail both of those equations). That makes my technique very quick.

Answer 12

eh @Luis_Rivera maybe, I haven't done brilliant in a while... it's mostly just knowing the multinomial theorem. Permutations of the variables and all that.