# solve gives 1, 2, or 3 answers depending if one value in my equation is a real, rational, or integer

Any idea why solve() is giving me different answers depending on what the value of 'g' is in my equations (see code below)? If g is a variable, an integer, or 1.0 then I get the two correct solutions (i.e. +&- a sqrt). If g is 1/4 or 1.1 I get an incomplete answer which I can get the two correct solutions by doing a solve on the solved solution. If g is a real whole number like 2,3 then I get 3 solutions: the two correct ones and a spurious one. I only discovered this behaviour when I tried `eqs=[0==f(x).diff(x)(x=0),0==f(x)(x=0)]; solve(eqs,T_1,x_0)`

and sage hanged just calculating forever. Is there any general rules about how I should enter variables in equations for use with solve()? I want to know beforehand if something like `/4`

will cause problems in which case I could use `v*`

instead and later, after using solve(), substitute with `.subs(v=1/4)`

?

```
reset()
forget()
var('T_1,T_2,x_0,h,d,g')
def f(x):
return B_0 + B_1 * (x - x_0) + B_2 * sqrt((x - x_0) ^ 2 + g*d)
B_0 = T_2 * (x_0 - h)
B_1 = (T_1 + T_2) / 2
B_2 = (T_2 - T_1) / 2
def check_solutions(gval):
print("g = {0}".format(gval))
eqs=[0==f(x).diff(x)(x=0,g=gval),0==f(x)(x=0,g=gval)]
print("******eqs:")
print(eqs)
sol1 = solve(eqs[0],T_1)
print("******Solution for T_1:")
print(sol1)
print("******sub T_1 solution into 2nd equation:")
print(eqs[1].subs(sol1[0]))
assume(h>0)
sol2 = solve(eqs[1].subs(sol1[0]),x_0)
print("******Solutions for x_0")
print(sol2)
print("******Solve again for x_0")
print([solve(sol2[i],x_0) for i in range(len(sol2))])
print("*****************************************")
gvals=[g,1,1.0,2.0,2,1/4,1.1]
for i in range(len(gvals)):
check_solutions(gvals[i])
```

running this I get:

```
g = g
******eqs:
[0 == 1/2*(T_1 - T_2)*x_0/sqrt(d*g + x_0^2) + 1/2*T_1 + 1/2*T_2, 0 == -(h - x_0)*T_2 - 1/2*(T_1 + T_2)*x_0 - 1/2*(T_1 - T_2)*sqrt(d*g + x_0^2)]
******Solution for T_1:
[
T_1 == (x_0 - sqrt(d*g + x_0^2))*T_2/(x_0 + sqrt(d*g + x_0^2))
]
******sub T_1 solution into 2nd equation:
0 == -(h - x_0)*T_2 - 1/2*((x_0 - sqrt(d*g + x_0^2))*T_2/(x_0 + sqrt(d*g + x_0^2)) + T_2)*x_0 - 1/2*((x_0 - sqrt(d*g + x_0^2))*T_2/(x_0 + sqrt(d*g + x_0^2)) - T_2)*sqrt(d*g + x_0^2)
******Solutions for x_0
[
x_0 == -sqrt(-d*g + h^2),
x_0 == sqrt(-d*g + h^2)
]
******Solve again for x_0
[[x_0 == -sqrt(-d*g + h^2)], [x_0 == sqrt(-d*g + h^2)]]
*****************************************
g = 1
******eqs:
[0 == 1/2*(T_1 - T_2)*x_0/sqrt(x_0^2 + d) + 1/2*T_1 + 1/2*T_2, 0 == -(h - x_0)*T_2 - 1/2*(T_1 + T_2)*x_0 - 1/2*(T_1 - T_2)*sqrt(x_0^2 + d)]
******Solution for T_1:
[
T_1 == (x_0 - sqrt(x_0 ...
```

This is a really good question, and likely has something to do with how we/Maxima convert certain constants - at times it replaces certain numbers by 'equivalent' fractions, I think (? or vice versa) which has caused related issues. If you can come up with a smaller example, or find the most distilled pieces of your example, that will make it that much easier to help you for us.

@kcrisman here's a smaller example that starts misbehaving for reals: reset() forget() var('h,x,g') assume (h>0) print(solve(sqrt(x+g)==h,x)[0]) print(solve(sqrt(x+1)==h,x)[0]) print(solve(sqrt(x+2.0)==h,x)[0]) print(solve(sqrt(x+1/4)==h,x)[0]) print(solve(sqrt(x+2.5)==h,x)[0]) myrr.<rr> = PolynomialRing(RR) print(solve(sqrt(x+rr)==h,x)[0]) which gives: x == h^2 - g x == h^2 - 1 x == h^2 - 2 1/2*sqrt(4*x + 1) == h 1/2*sqrt(2*x + 5)*sqrt(2) == h and a traceback error on the last example with rr (not enough space in comment to show). Other polynomial rings (CC,ZZ,QQ) gave the correct answer.

Also, in my original question if I replace sol2 = solve(eqs[1].subs(sol1[0]),x_0) with sol2 = solve(eqs[1].subs(sol1[0]).expand(),x_0), i.e. expand before solving, I get x0 = some function of x0 which is wrong.

In addition to my first comment to @kcrisman for the case where g = 1/4 if I run solve() on the incomplete answer that solve() originally gave me then I get the correct answer: solve(1/2*sqrt(4*x + 1) == h,x) gives: x == h^2 - 1/4 Why doesn't solve() give me that answer in the first place?