Why do I get more real solutions then complex solutions for a equation?

I tried to solve an equation using the MATLAB symbolic toolbox:

syms x;
solve(abs(x+1)+abs(5*x-2)==6,x);

The returned result is

ans=-0.75;

If I only allow real solutions by typing

syms x real
solve(abs(x+1)+abs(5*x-2)==6,x)

I get two real solutions

ans =

    7/6
   -3/4

Why is MATLAB unable to find the second solution (7/6) when solving the equation for complex x?

[Adam]

I don't know nor have the Symbolic Math Toolbox, but surely even the 0.75 answer that it does return is not a solution to that equation either is it?

[Andrew Reibold]

The solution is -0.75, not .075

I think he just made a typo the first time, he has it in the second solution.

I like this question though, I also am interested in the answer.

[Adam]

Yeah, I thought it might be, it just looked as though it would have been a copy and pasted result!

[John D'Errico]

My answer should resolve your questions, I hope.

Answer 1

[John D'Errico]

I would point out it is important to read what the toolbox tells you.

syms x;
solve(abs(x+1)+abs(5*x-2)==6,x)
Warning: Cannot solve symbolically. Returning a numeric approximation instead. 
> In solve at 306 
ans =
-0.75

So it had problems solving the problem, so it used a numerical solver instead! READ WHAT IT TELLS YOU! It did not find a symbolic solution, as if X is allowed to be complex, there would be insufficient information to solve the problem.

Think of it like this, if x is complex, it is really a two variable problem, such that

x = x_r + i*x_i

So a real and imaginary part, where both of those variables would be take on real values only. But you have only one equation, and the output of the function you have posed is always real, since it has those abs functions in it. So essentially, you have one equation, with two unknowns. My guess is there may be an infinite number of solutions if you pose this problem in the complex plane.

Lets see how this works if we expand the problem as the symbolic toolbox might implicitly do internally. By the way, the square roots below will always be the principle square root, so the positive branch.

real(sqrt((x_r + 1)^2 + x_i^2) + sqrt((5*x_r - 2)^2 + (5*x_i)^2))) == 6
imag(sqrt((x_r + 1)^2 + x_i^2) + sqrt((5*x_r - 2)^2 + (5*x_i)^2))) == 0

See that the complex equality is implicitly two equality statements, but the second one, where we take the imaginary part is ALWAYS trivially zero, so it imparts no information about the solution. So we are left with this single equation in two real unknowns, (x_r,x_i).

sqrt((x_r + 1)^2 + x_i^2) + sqrt((5*x_r - 2)^2 + (5*x_i)^2)) == 6

The solution in general to such a problem is usually a curved path in the (x_r,x_i) plane, thus, the complex plane. This is why I suggest there are infinitely many solutions.

We can see the solution locus easily enough, letting contour do the hard work for us.

[x_r,x_i] = meshgrid(-3:.01:3);
contour(x_r,x_i,sqrt((x_r + 1).^2 + x_i.^2) + sqrt((5*x_r - 2).^2 + 5*x_i).^2),[6 6])
grid on
axis square
axis equal

As you can see, there are infinitely many solutions if we look in the complex plane. The two real solutions (x_r, such that x_i==0) appear to be -0.75, and 1.1666667, which matches what solve told you for real solutions.

The problem is indeed much simpler when x is purely real of course, since then it reduces to

abs(x_r + 1) + abs(5*x_r - 2) == 6

We can see the solutions to this graphically. I'll add a reference line at y==6 to indicate the solutions where our function crosses that line.

ezplot(@(x_r) abs(x_r + 1) + abs(5*x_r - 2))
grid on
hold on
plot([-6 6],[6 6],'r-')

The two points of intersection correspond to the real solutions returned by solve.

Of course, solve did return a -0.75 solution, which is indeed one of the valid solutions that solve returned when x was constrained to be purely real. This was generated by a numerical solver, and numerical solvers cannot be expected to return all the solutions to a problem.

[Benjamin]

Thank you for your great explanation!

But this brings me to my next question: Why is MATLAB not displaying this quite useful warning to me?

I only get the numerical approximation of the solution without the warning:

>> syms x
>> solve(abs(x+1)+abs(5*x-2)==6,x)

ans =

-0.75

I'm using MATLAB R2014a and usually warnings appear when trying to do things where I should be warned:

>> A=[1 0 0;0 1 0;0 1 0]

A =

       1     0     0
       0     1     0
       0     1     0

>> A^-1
Warning: Matrix is singular to working precision. 

ans =

     Inf   Inf   Inf
     Inf   Inf   Inf
     Inf   Inf   Inf

[Sean de Wolski]

Do you have warnings turned off? It does for me.

syms  x
solve(abs(x+1)+abs(5*x-2)==6,x)
Warning: Cannot solve symbolically. Returning a numeric
approximation instead. 
> In solve at 306

Maybe:

doc warning

[John D'Errico]

My guess is warnings must be turned off. (If I turn them off, I get the same non-informatory result as you did.) They can get that way, though I'm not sure why that happened for you. Try this:

warning('on')

The retry the solve call and see if a warning happens.