How to obtain all the solutions of quadratic equations with multiple variables?

I am trying to solve quadratic equations with multiple variables. For example, I want to solve

a^2+b^2 = 1.25;
c^2+d^2 = 0.58;
a*c+b*d = 0.55;
a*d-b*c = 0.65;

The code I wrote is as follows.

>> syms a b c d
>> eqns = [a^2+b^2 == 1.25, c^2+d^2 == 0.58, a*c+b*d == 0.55, a*d-b*c == 0.65];
[sola,solb,solc,sold] = solve(eqns,[a b c d])
sola =

 -(11*2^(1/2)*29^(1/2))/116
  (11*2^(1/2)*29^(1/2))/116 
solb =

(13*2^(1/2)*29^(1/2))/116
-(13*2^(1/2)*29^(1/2))/116
solc = 

 -(2^(1/2)*29^(1/2))/10
  (2^(1/2)*29^(1/2))/10
sold =

 0
 0

Matlab provides two sets of solutions, but they are not all the sets of solutions. At least [1, -0.5, 0.7, 0.3] is another set of solution. Why Matlab doesn't provide all the solutions when using 'solve' function? How to obtain all the solution?

Can anyone help? Thanks in advance.

Answer 1

[Bruno Luong]

I have impression the 4 equations are not independent, and that might confuse MATLAB.

For any 2x2 rotation matrix

R = [cos(theta), -sin(theta); 
     sin(theta) cos(theta)}

I believe if (a,b,c,d) a solution, if I define

abp = R*[a;b];
cdp = R*[c;d];

then

(abp(1),abp(2),cdp(1),cdp(2))

is another solution for any theta not a multiple of 2*pi.

So actually there are an infinity of solutions. Of course MATLAB won't be able to list them. But pretending cardinal of all solutions is 2 is obviously very wrong.

And can also prove that any solution is rotated (as described above) from another.

[Huanyu Zuo]

Awesome! You find the general solution for the equations.

I also found Matlab can provide the complete set of solutions for the equations.

>> syms a b c d
>> eqns=[a^2+b^2 == 1.25, c^2+d^2 == 0.58, a*c+b*d == 0.55, a*d-b*c == 0.65];
>> [sola,solb,solc,sold,Para,Cond] = solve(eqns,[a b c d],'ReturnConditions', true)

sola =

   (65*z)/58 - (11*2^(1/2)*(29 - 50*z^2)^(1/2))/116
   (65*z)/58 + (11*2^(1/2)*(29 - 50*z^2)^(1/2))/116

solb =

   (55*z)/58 + (13*2^(1/2)*(29 - 50*z^2)^(1/2))/116
   (55*z)/58 - (13*2^(1/2)*(29 - 50*z^2)^(1/2))/116

solc =

   -(2^(1/2)*(29 - 50*z^2)^(1/2))/10
    (2^(1/2)*(29 - 50*z^2)^(1/2))/10

sold =

   z
   z

Para =

z

Cond =

   TRUE
   TRUE

The condition is 'true', but I am not sure if it means 'z' can be any values.

[Huanyu Zuo]

I tried and it seems that 'z' can be any values indeed, even for a complex value.

[Huanyu Zuo]

I accepted your answer, though you did not tell me how to obtain all the solutions using Matlab. The way you think is very impressive and you provide me a method to deal with similar questions. Thank you!

[Bruno Luong]

As I don't own SYMBOLIC TBx I can't help you to deal with MATLAB, but at least with some though. Thanks, glad to help.

Answer 2

[madhan ravi]

As you can see you got all the solutions , each variable has two solutions. So I don’t see anything wrong in it.

[Bruno Luong]

he claims "At least [1, -0.5, 0.7, 0.3] is another set of solution."

[madhan ravi]

I don't understand what it means , he wants one minimum set of solution is it ? can you make it clear because I don't understand

[Bruno Luong]

It means MATLAB is not able to give ALL solution

[madhan ravi]

but as you can see he got all the solutions as I mentioned in my answer because there arises in equation with the power two for instance the values correlate if double(sola) except sold is 0 instead of 0.3

[Bruno Luong]

NO because the solution he gives in NOT in this set (I checked and it looks likes solution), MATLAB tells d must be 0, but his solution has d=0.3, so MATLAB miss this solution.

[madhan ravi]

yes exactly , don't know why Matlab misses the last solution

[Bruno Luong]

There might be even more that MATLAB misses. A bug?

[madhan ravi]

No idea Bruno might be or the precision isn’t accurate?

[Bruno Luong]

Precision in symbolic calculation? Nah...

[madhan ravi]

then might be as you suggested

[Huanyu Zuo]

Yeah, Matlab misses more indeed. Let a be 0, we can get another two sets of solutions. Same thing happens if we let b or c be 0.

[madhan ravi]

can you tell where you got those suggestions so that we could also check it

[Bruno Luong]

Just assign a = 0 and solve for b,c,d, etc...

[Huanyu Zuo]

Yeah, just as Bruno said. For example,

>> b=0;
>> syms a c d
>> eqns=[a^2+b^2==1.25,c^2+d^2==0.58,a*c+b*d==0.55,a*d-b*c==0.65];
>> [sola,solc,sold]=solve(eqns,[a c d])

sola =

   -5^(1/2)/2
    5^(1/2)/2

solc =

   -(11*5^(1/2))/50
    (11*5^(1/2))/50

sold =

   -(13*5^(1/2))/50
    (13*5^(1/2))/50

They are the new sets of solutions.

[madhan ravi]

yea I get it know