Ok. It looks like your real question is where does the spurious root arise? And, since you are using an old enough release of MATLAB that it will only run on a computer old enough that it needs a crank on the side to start it, I can't run that release to compare. But we can see from whence the spurious solution arises.
First, we need to recognize what happens with a square root. The sqrt function actually has TWO branches, even though people only tend to think of one. So if u=sqrt(x), then so is -u a valid square root. Writing your relation as...
x - sqrt(x+1) - sqrt(x-1) == 0
we really need to think of it as 4 possible problems, thus these three others.
x + sqrt(x+1) + sqrt(x-1) == 0
x + sqrt(x+1) - sqrt(x-1) == 0
x - sqrt(x+1) + sqrt(x-1) == 0
So lets plot all 4 relations on one set of axes. In the plot, sqrt will always takes the positive branch only.
ezplot(@(x) x - sqrt(x+1) - sqrt(x-1),[-1,5])
hold on
ezplot(@(x) x + sqrt(x+1) + sqrt(x-1),[-1,5])
ezplot(@(x) x + sqrt(x+1) - sqrt(x-1),[-1,5])
ezplot(@(x) x - sqrt(x+1) + sqrt(x-1),[-1,5])
refline([0,0])
As you should see, there are 4 different curves there. distinguished by the 4 colors plotted. As well, there are TWO solutions, since both the cyan and blue curves cross the reference line at y==0.
So truly, there are two distinct and fully valid real solutions, as long as you agree that the square root function has two branches.
Your version of solve found both of them, one of which is apparently at 1.11508. It arises as the solution to this variation of your problem (I am using the R2018b version):
syms x
vpa(solve(x-sqrt(x+1)+sqrt(x-1)))
ans =
1.1150879946798484383326671302376
So my version of solve seems to be looking at the positive branch of the sqrt function only.
It is simple enough to solve the problem in a way that generates the spurious solution too.
x = sqrt(x-1) + sqrt(x+1)
Square both sides, to get
x^2 = (x-1) + (x+1) + 2*sqrt((x-1)*(x+1))
Collect, then square again.
(x^2 - 2*x)^2 = 4*(x-1)*(x+1)
See that by squaring things, we resolve those +/- ambiguities. But at the same time, we may introduce spurious solutions, since sqrt as a function tends to imply only the positive square root to many people who want to ignore the negative branch.
That 4th degree polynomial has 4 roots, some of which may be complex. They are rather messy to write in full form, as you might expect.
R = solve((x^2 - 2*x)^2 == 4*(x-1)*(x+1),'maxdegree',4)
R =
1 - (3^(1/2)*(24*3^(1/2)*((32*3^(1/2)*23^(1/2))/9 + 32)^(1/3)*(12*((32*3^(1/2)*23^(1/2))/9 + 32)^(1/3) + 3*((32*3^(1/2)*23^(1/2))/9 + 32)^(2/3) + 16)^(1/2) - 16*3^(1/2)*(12*((32*3^(1/2)*23^(1/2))/9 + 32)^(1/3) + 3*((32*3^(1/2)*23^(1/2))/9 + 32)^(2/3) + 16)^(1/2) - 3*3^(1/2)*((32*3^(1/2)*23^(1/2))/9 + 32)^(2/3)*(12*((32*3^(1/2)*23^(1/2))/9 + 32)^(1/3) + 3*((32*3^(1/2)*23^(1/2))/9 + 32)^(2/3) + 16)^(1/2) - 64*3^(1/2)*6^(1/2)*(3^(1/2)*23^(1/2) + 9)^(1/2))^(1/2))/(6*((32*3^(1/2)*23^(1/2))/9 + 32)^(1/6)*(36*((32*3^(1/2)*23^(1/2))/9 + 32)^(1/3) + 9*((32*3^(1/2)*23^(1/2))/9 + 32)^(2/3) + 48)^(1/4)) - (3^(1/2)*(12*((32*3^(1/2)*23^(1/2))/9 + 32)^(1/3) + 3*((32*3^(1/2)*23^(1/2))/9 + 32)^(2/3) + 16)^(1/2))/(6*((32*3^(1/2)*23^(1/2))/9 + 32)^(1/6))
(3^(1/2)*(24*3^(1/2)*((32*3^(1/2)*23^(1/2))/9 + 32)^(1/3)*(12*((32*3^(1/2)*23^(1/2))/9 + 32)^(1/3) + 3*((32*3^(1/2)*23^(1/2))/9 + 32)^(2/3) + 16)^(1/2) - 16*3^(1/2)*(12*((32*3^(1/2)*23^(1/2))/9 + 32)^(1/3) + 3*((32*3^(1/2)*23^(1/2))/9 + 32)^(2/3) + 16)^(1/2) - 3*3^(1/2)*((32*3^(1/2)*23^(1/2))/9 + 32)^(2/3)*(12*((32*3^(1/2)*23^(1/2))/9 + 32)^(1/3) + 3*((32*3^(1/2)*23^(1/2))/9 + 32)^(2/3) + 16)^(1/2) - 64*3^(1/2)*6^(1/2)*(3^(1/2)*23^(1/2) + 9)^(1/2))^(1/2))/(6*((32*3^(1/2)*23^(1/2))/9 + 32)^(1/6)*(36*((32*3^(1/2)*23^(1/2))/9 + 32)^(1/3) + 9*((32*3^(1/2)*23^(1/2))/9 + 32)^(2/3) + 48)^(1/4)) - (3^(1/2)*(12*((32*3^(1/2)*23^(1/2))/9 + 32)^(1/3) + 3*((32*3^(1/2)*23^(1/2))/9 + 32)^(2/3) + 16)^(1/2))/(6*((32*3^(1/2)*23^(1/2))/9 + 32)^(1/6)) + 1
(3^(1/2)*(12*((32*3^(1/2)*23^(1/2))/9 + 32)^(1/3) + 3*((32*3^(1/2)*23^(1/2))/9 + 32)^(2/3) + 16)^(1/2))/(6*((32*3^(1/2)*23^(1/2))/9 + 32)^(1/6)) - (3^(1/2)*(24*3^(1/2)*((32*3^(1/2)*23^(1/2))/9 + 32)^(1/3)*(12*((32*3^(1/2)*23^(1/2))/9 + 32)^(1/3) + 3*((32*3^(1/2)*23^(1/2))/9 + 32)^(2/3) + 16)^(1/2) - 16*3^(1/2)*(12*((32*3^(1/2)*23^(1/2))/9 + 32)^(1/3) + 3*((32*3^(1/2)*23^(1/2))/9 + 32)^(2/3) + 16)^(1/2) - 3*3^(1/2)*((32*3^(1/2)*23^(1/2))/9 + 32)^(2/3)*(12*((32*3^(1/2)*23^(1/2))/9 + 32)^(1/3) + 3*((32*3^(1/2)*23^(1/2))/9 + 32)^(2/3) + 16)^(1/2) + 64*3^(1/2)*6^(1/2)*(3^(1/2)*23^(1/2) + 9)^(1/2))^(1/2))/(6*((32*3^(1/2)*23^(1/2))/9 + 32)^(1/6)*(36*((32*3^(1/2)*23^(1/2))/9 + 32)^(1/3) + 9*((32*3^(1/2)*23^(1/2))/9 + 32)^(2/3) + 48)^(1/4)) + 1
(3^(1/2)*(12*((32*3^(1/2)*23^(1/2))/9 + 32)^(1/3) + 3*((32*3^(1/2)*23^(1/2))/9 + 32)^(2/3) + 16)^(1/2))/(6*((32*3^(1/2)*23^(1/2))/9 + 32)^(1/6)) + (3^(1/2)*(24*3^(1/2)*((32*3^(1/2)*23^(1/2))/9 + 32)^(1/3)*(12*((32*3^(1/2)*23^(1/2))/9 + 32)^(1/3) + 3*((32*3^(1/2)*23^(1/2))/9 + 32)^(2/3) + 16)^(1/2) - 16*3^(1/2)*(12*((32*3^(1/2)*23^(1/2))/9 + 32)^(1/3) + 3*((32*3^(1/2)*23^(1/2))/9 + 32)^(2/3) + 16)^(1/2) - 3*3^(1/2)*((32*3^(1/2)*23^(1/2))/9 + 32)^(2/3)*(12*((32*3^(1/2)*23^(1/2))/9 + 32)^(1/3) + 3*((32*3^(1/2)*23^(1/2))/9 + 32)^(2/3) + 16)^(1/2) + 64*3^(1/2)*6^(1/2)*(3^(1/2)*23^(1/2) + 9)^(1/2))^(1/2))/(6*((32*3^(1/2)*23^(1/2))/9 + 32)^(1/6)*(36*((32*3^(1/2)*23^(1/2))/9 + 32)^(1/3) + 9*((32*3^(1/2)*23^(1/2))/9 + 32)^(2/3) + 48)^(1/4)) + 1
We can resolve them into decimal form using vpa.
vpa(R)
ans =
- 0.52470257992985177015834395726155 - 0.79777765915011117379630925220011i
- 0.52470257992985177015834395726155 + 0.79777765915011117379630925220011i
1.1150879946798484383326671302376
3.9343171651798551019840207842855
So there are two real solutions to the problem once converted into an associated 4th degree polynomial. But only one of them is a solution to the original problem as posed, IF you take only the positive branch of the sqrt.
In the end, you should see that solve did not return an inaccurate solution, but just a surprising one, because you did not take into account the two branches of a sqrt. Is it a spurious solution? I supose you can argue that either way.
