Is Mathematica able to solve expressions such as

In summary, Mathematica is a computational software program that uses advanced algorithms and symbolic computation to solve a wide range of mathematical expressions and equations in various fields. It can handle large expressions and datasets with efficiency and accuracy, and also has built-in functions for data analysis and visualization.
  • #1
binbagsss
1,259
11
(c (qz + (-1 + q) (-1 + (1 + qz)^0.5)))/(h q^2 (1 + z)^2) = d, And to solve for q

New to the programme, thanks a lot !
 
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  • #2
I think this answers your question

In[1]:= Reduce[c (q*z+(-1+q) (-1+Sqrt[1+q*z]))==d*(h*q^2 (1+z)^2),q]

From In[1]:= Reduce::useq: The answer found by Reduce contains unsolved equation(s) A likely reason for this is that the solution set depends on branch cuts of Mathematica functions.

Out[1]= (z == -1 && c == 0) || (z == 0 && h == 0 && c == 0) || (c != 0 &&
z == -1 && q == 0) || c != 0 && z == -1 && q == (-1 - Sqrt[5])/2) || (c*(-1
+ q) != 0 && z == 0 && h == 0) || (d*h*(1 + z) != 0 && -c != 0 && q == 0) || (z
+ z^2 != 0 && h == 0 && c == 0) || (h*z + h*z^2 != 0 && d == 0 && c == 0) ||
(d*h*z + d*h*z^2 != 0 && c == 0 && q == 0) || (c != 0 && z == 0 && h == 0 && q
== 1) || (h != 0 && z == 0 && d == 0 && c == 0) || (d*h != 0 && z == 0 && c == 0
&& q == 0) || (c*(-1 + q) != 0 && h != 0 && z == 0 && d == 0) || (z != 0 && d*h
+ 2*d*h*z + d*h*z^2 != 0 && c == (d*h + 2*d*h*z + d*h*z^2)/z && q == 1) || (c*z
!= 0 && 1 + z != 0 && h == 0 && q == 0) || (c*z != 0 && h + h*z != 0 && d == 0
&& q == 0) || (d*h*(1 + z) != 0 && c*(-1 + Root[c^2*z - 2*c*d*h*#1 -
4*c*d*h*z*#1 + c^2*z^2*#1 - 2*c*d*h*z^2*#1 + 2*c*d*h* #1^2 - c^2*z*#1^2 +
2*c*d*h*z*#1^2 - 2*c*d*h*z^2*#1^2 - 2*c*d*h*z^3*#1^2 + d^2*h^2*#1^3 +
4*d^2*h^2*z* #1^3 + 6*d^2*h^2*z^2*#1^3 + 4*d^2*h^2*z^3*#1^3 + d^2*h^2*z^4*#1^3 &
, 1]) != 0 && 0 == (-c + c*Root[c^2*z - 2*c*d*h*#1 - 4*c*d*h*z*#1 + c^2*z^2*#1 -
2*c*d*h*z^2*#1 + 2*c*d*h* #1^2 - c^2*z*#1^2 + 2*c*d*h*z*#1^2 - 2*c*d*h*z^2* #1^2
- 2*c*d*h*z^3*#1^2 + d^2*h^2*#1^3 + 4*d^2*h^2*z* #1^3 + 6*d^2*h^2*z^2*#1^3 +
4*d^2*h^2*z^3*#1^3 + d^2*h^2*z^4*#1^3 & , 1] - c*z*Root[c^2*z - 2*c*d*h*#1 -
4*c*d*h*z*#1 + c^2*z^2*#1 - 2*c*d*h*z^2*#1 + 2*c*d*h* #1^2 - c^2*z*#1^2 +
2*c*d*h*z*#1^2 - 2*c*d*h*z^2* #1^2 - 2*c*d*h*z^3*#1^2 + d^2*h^2*#1^3 +
4*d^2*h^2*z* #1^3 + 6*d^2*h^2*z^2*#1^3 + 4*d^2*h^2*z^3*#1^3 + d^2*h^2*z^4*#1^3 &
, 1] + d*h*Root[c^2*z - 2*c*d*h*#1 - 4*c*d*h*z*#1 + c^2*z^2*#1 - 2*c*d*h*z^2*#1
+ 2*c*d*h* #1^2 - c^2*z*#1^2 + 2*c*d*h*z*#1^2 - 2*c*d*h* z^2*#1^2 - 2*c*d*h*z^3*
#1^2 + d^2*h^2*#1^3 + 4*d^2*h^2*z*#1^3 + 6*d^2*h^2*z^2*#1^3 + 4*d^2*h^2*z^3*#1^3
+ d^2*h^2*z^4*#1^3 & , 1]^2 + 2*d*h*z* Root[c^2*z - 2*c*d*h*#1 - 4*c*d*h*z*#1 +
c^2*z^2*#1 - 2*c*d*h*z^2*#1 + 2*c*d*h* #1^2 - c^2*z*#1^2 + 2*c*d*h*z*#1^2 -
2*c*d*h* z^2*#1^2 - 2*c*d*h*z^3* #1^2 + d^2*h^2*#1^3 + 4*d^2*h^2*z*#1^3 +
6*d^2*h^2*z^2*#1^3 + 4*d^2*h^2*z^3*#1^3 + d^2*h^2*z^4*#1^3 & , 1]^2 + d*h*z^2*
Root[c^2*z - 2*c*d*h*#1 - 4*c*d*h*z*#1 + c^2*z^2*#1 - 2*c*d*h*z^2*#1 + 2*c*d*h*
#1^2 - c^2*z*#1^2 + 2*c*d*h*z*#1^2 - 2*c*d*h* z^2*#1^2 - 2*c*d*h*z^3* #1^2 +
d^2*h^2*#1^3 + 4*d^2*h^2*z*#1^3 + 6*d^2*h^2*z^2*#1^3 + 4*d^2*h^2*z^3*#1^3 +
d^2*h^2*z^4*#1^3 & , 1]^2 + c*Sqrt[1 + z*Root[ c^2*z - 2*c*d*h*#1 - 4*c*d*h*z*#1
+ c^2*z^2* #1 - 2*c*d*h*z^2*#1 + 2*c*d*h*#1^2 - c^2*z* #1^2 + 2*c*d*h*z*#1^2 -
2*c*d*h*z^2*#1^2 - 2*c*d*h*z^3*#1^2 + d^2*h^2*#1^3 + 4*d^2*h^2* z*#1^3 +
6*d^2*h^2*z^2* #1^3 + 4*d^2*h^2*z^3* #1^3 + d^2*h^2*z^4* #1^3 & , 1]] -
c*Root[c^2*z - 2*c*d*h*#1 - 4*c*d*h*z*#1 + c^2*z^2*#1 - 2*c*d*h*z^2*#1 +
2*c*d*h* #1^2 - c^2*z*#1^2 + 2*c*d*h*z*#1^2 - 2*c*d*h*z^2* #1^2 -
2*c*d*h*z^3*#1^2 + d^2*h^2*#1^3 + 4*d^2*h^2*z* #1^3 + 6*d^2*h^2*z^2*#1^3 +
4*d^2*h^2*z^3*#1^3 + d^2*h^2*z^4*#1^3 & , 1]* Sqrt[1 + z*Root[ c^2*z -
2*c*d*h*#1 - 4*c*d*h*z*#1 + c^2*z^2* #1 - 2*c*d*h*z^2*#1 + 2*c*d*h*#1^2 - c^2*z*
#1^2 + 2*c*d*h*z*#1^2 - 2*c*d*h*z^2*#1^2 - 2*c*d*h*z^3*#1^2 + d^2*h^2*#1^3 +
4*d^2*h^2* z*#1^3 + 6*d^2*h^2*z^2* #1^3 + 4*d^2*h^2*z^3* #1^3 + d^2*h^2*z^4*
#1^3 & , 1]])/ (c*(-1 + Root[c^2*z - 2*c*d*h* #1 - 4*c*d*h*z*#1 + c^2*z^2*#1 -
2*c*d*h*z^2* #1 + 2*c*d*h*#1^2 - c^2*z*#1^2 + 2*c*d*h*z* #1^2 - 2*c*d*h*z^2*#1^2
- 2*c*d*h*z^3*#1^2 + d^2*h^2*#1^3 + 4*d^2*h^2*z* #1^3 + 6*d^2*h^2*z^2*#1^3 +
4*d^2*h^2*z^3*#1^3 + d^2*h^2*z^4*#1^3 & , 1])) && q == Root[c^2*z - 2*c*d*h*#1 -
4*c*d*h*z*#1 + c^2*z^2*#1 - 2*c*d*h*z^2*#1 + 2*c*d*h*#1^2 - c^2*z*#1^2 +
2*c*d*h*z*#1^2 - 2*c*d*h*z^2*#1^2 - 2*c*d*h*z^3* #1^2 + d^2*h^2*#1^3 +
4*d^2*h^2*z*#1^3 + 6*d^2*h^2* z^2*#1^3 + 4*d^2*h^2*z^3* #1^3 + d^2*h^2*z^4*#1^3
& , 1]) || (d*h*(1 + z) != 0 && c*(-1 + Root[c^2*z - 2*c*d*h*#1 - 4*c*d*h*z*#1 +
c^2*z^2*#1 - 2*c*d*h*z^2*#1 + 2*c*d*h* #1^2 - c^2*z*#1^2 + 2*c*d*h*z*#1^2 -
2*c*d*h*z^2* #1^2 - 2*c*d*h*z^3*#1^2 + d^2*h^2*#1^3 + 4*d^2*h^2*z* #1^3 +
6*d^2*h^2*z^2*#1^3 + 4*d^2*h^2*z^3*#1^3 + d^2*h^2*z^4*#1^3 & , 2]) != 0 && 0 ==
(-c + c*Root[c^2*z - 2*c*d*h*#1 - 4*c*d*h*z*#1 + c^2*z^2*#1 - 2*c*d*h*z^2*#1 +
2*c*d*h* #1^2 - c^2*z*#1^2 + 2*c*d*h*z*#1^2 - 2*c*d*h*z^2* #1^2 -
2*c*d*h*z^3*#1^2 + d^2*h^2*#1^3 + 4*d^2*h^2*z* #1^3 + 6*d^2*h^2*z^2*#1^3 +
4*d^2*h^2*z^3*#1^3 + d^2*h^2*z^4*#1^3 & , 2] - c*z*Root[c^2*z - 2*c*d*h*#1 -
4*c*d*h*z*#1 + c^2*z^2*#1 - 2*c*d*h*z^2*#1 + 2*c*d*h* #1^2 - c^2*z*#1^2 +
2*c*d*h*z*#1^2 - 2*c*d*h*z^2* #1^2 - 2*c*d*h*z^3*#1^2 + d^2*h^2*#1^3 +
4*d^2*h^2*z* #1^3 + 6*d^2*h^2*z^2*#1^3 + 4*d^2*h^2*z^3*#1^3 + d^2*h^2*z^4*#1^3 &
, 2] + d*h*Root[c^2*z - 2*c*d*h*#1 - 4*c*d*h*z*#1 + c^2*z^2*#1 - 2*c*d*h*z^2*#1
+ 2*c*d*h* #1^2 - c^2*z*#1^2 + 2*c*d*h*z*#1^2 - 2*c*d*h* z^2*#1^2 - 2*c*d*h*z^3*
#1^2 + d^2*h^2*#1^3 + 4*d^2*h^2*z*#1^3 + 6*d^2*h^2*z^2*#1^3 + 4*d^2*h^2*z^3*#1^3
+ d^2*h^2*z^4*#1^3 & , 2]^2 + 2*d*h*z* Root[c^2*z - 2*c*d*h*#1 - 4*c*d*h*z*#1 +
c^2*z^2*#1 - 2*c*d*h*z^2*#1 + 2*c*d*h* #1^2 - c^2*z*#1^2 + 2*c*d*h*z*#1^2 -
2*c*d*h* z^2*#1^2 - 2*c*d*h*z^3* #1^2 + d^2*h^2*#1^3 + 4*d^2*h^2*z*#1^3 +
6*d^2*h^2*z^2*#1^3 + 4*d^2*h^2*z^3*#1^3 + d^2*h^2*z^4*#1^3 & , 2]^2 + d*h*z^2*
Root[c^2*z - 2*c*d*h*#1 - 4*c*d*h*z*#1 + c^2*z^2*#1 - 2*c*d*h*z^2*#1 + 2*c*d*h*
#1^2 - c^2*z*#1^2 + 2*c*d*h*z*#1^2 - 2*c*d*h* z^2*#1^2 - 2*c*d*h*z^3* #1^2 +
d^2*h^2*#1^3 + 4*d^2*h^2*z*#1^3 + 6*d^2*h^2*z^2*#1^3 + 4*d^2*h^2*z^3*#1^3 +
d^2*h^2*z^4*#1^3 & , 2]^2 + c*Sqrt[1 + z*Root[ c^2*z - 2*c*d*h*#1 - 4*c*d*h*z*#1
+ c^2*z^2* #1 - 2*c*d*h*z^2*#1 + 2*c*d*h*#1^2 - c^2*z* #1^2 + 2*c*d*h*z*#1^2 -
2*c*d*h*z^2*#1^2 - 2*c*d*h*z^3*#1^2 + d^2*h^2*#1^3 + 4*d^2*h^2* z*#1^3 +
6*d^2*h^2*z^2* #1^3 + 4*d^2*h^2*z^3* #1^3 + d^2*h^2*z^4* #1^3 & , 2]] -
c*Root[c^2*z - 2*c*d*h*#1 - 4*c*d*h*z*#1 + c^2*z^2*#1 - 2*c*d*h*z^2*#1 +
2*c*d*h* #1^2 - c^2*z*#1^2 + 2*c*d*h*z*#1^2 - 2*c*d*h*z^2* #1^2 -
2*c*d*h*z^3*#1^2 + d^2*h^2*#1^3 + 4*d^2*h^2*z* #1^3 + 6*d^2*h^2*z^2*#1^3 +
4*d^2*h^2*z^3*#1^3 + d^2*h^2*z^4*#1^3 & , 2]* Sqrt[1 + z*Root[ c^2*z -
2*c*d*h*#1 - 4*c*d*h*z*#1 + c^2*z^2* #1 - 2*c*d*h*z^2*#1 + 2*c*d*h*#1^2 - c^2*z*
#1^2 + 2*c*d*h*z*#1^2 - 2*c*d*h*z^2*#1^2 - 2*c*d*h*z^3*#1^2 + d^2*h^2*#1^3 +
4*d^2*h^2* z*#1^3 + 6*d^2*h^2*z^2* #1^3 + 4*d^2*h^2*z^3* #1^3 + d^2*h^2*z^4*
#1^3 & , 2]])/ (c*(-1 + Root[c^2*z - 2*c*d*h* #1 - 4*c*d*h*z*#1 + c^2*z^2*#1 -
2*c*d*h*z^2* #1 + 2*c*d*h*#1^2 - c^2*z*#1^2 + 2*c*d*h*z* #1^2 - 2*c*d*h*z^2*#1^2
- 2*c*d*h*z^3*#1^2 + d^2*h^2*#1^3 + 4*d^2*h^2*z* #1^3 + 6*d^2*h^2*z^2*#1^3 +
4*d^2*h^2*z^3*#1^3 + d^2*h^2*z^4*#1^3 & , 2])) && q == Root[c^2*z - 2*c*d*h*#1 -
4*c*d*h*z*#1 + c^2*z^2*#1 - 2*c*d*h*z^2*#1 + 2*c*d*h*#1^2 - c^2*z*#1^2 +
2*c*d*h*z*#1^2 - 2*c*d*h*z^2*#1^2 - 2*c*d*h*z^3* #1^2 + d^2*h^2*#1^3 +
4*d^2*h^2*z*#1^3 + 6*d^2*h^2* z^2*#1^3 + 4*d^2*h^2*z^3* #1^3 + d^2*h^2*z^4*#1^3
& , 2]) || (d*h*(1 + z) != 0 && c*(-1 + Root[c^2*z - 2*c*d*h*#1 - 4*c*d*h*z*#1 +
c^2*z^2*#1 - 2*c*d*h*z^2*#1 + 2*c*d*h* #1^2 - c^2*z*#1^2 + 2*c*d*h*z*#1^2 -
2*c*d*h*z^2* #1^2 - 2*c*d*h*z^3*#1^2 + d^2*h^2*#1^3 + 4*d^2*h^2*z* #1^3 +
6*d^2*h^2*z^2*#1^3 + 4*d^2*h^2*z^3*#1^3 + d^2*h^2*z^4*#1^3 & , 3]) != 0 && 0 ==
(-c + c*Root[c^2*z - 2*c*d*h*#1 - 4*c*d*h*z*#1 + c^2*z^2*#1 - 2*c*d*h*z^2*#1 +
2*c*d*h* #1^2 - c^2*z*#1^2 + 2*c*d*h*z*#1^2 - 2*c*d*h*z^2* #1^2 -
2*c*d*h*z^3*#1^2 + d^2*h^2*#1^3 + 4*d^2*h^2*z* #1^3 + 6*d^2*h^2*z^2*#1^3 +
4*d^2*h^2*z^3*#1^3 + d^2*h^2*z^4*#1^3 & , 3] - c*z*Root[c^2*z - 2*c*d*h*#1 -
4*c*d*h*z*#1 + c^2*z^2*#1 - 2*c*d*h*z^2*#1 + 2*c*d*h* #1^2 - c^2*z*#1^2 +
2*c*d*h*z*#1^2 - 2*c*d*h*z^2* #1^2 - 2*c*d*h*z^3*#1^2 + d^2*h^2*#1^3 +
4*d^2*h^2*z* #1^3 + 6*d^2*h^2*z^2*#1^3 + 4*d^2*h^2*z^3*#1^3 + d^2*h^2*z^4*#1^3 &
, 3] + d*h*Root[c^2*z - 2*c*d*h*#1 - 4*c*d*h*z*#1 + c^2*z^2*#1 - 2*c*d*h*z^2*#1
+ 2*c*d*h* #1^2 - c^2*z*#1^2 + 2*c*d*h*z*#1^2 - 2*c*d*h* z^2*#1^2 - 2*c*d*h*z^3*
#1^2 + d^2*h^2*#1^3 + 4*d^2*h^2*z*#1^3 + 6*d^2*h^2*z^2*#1^3 + 4*d^2*h^2*z^3*#1^3
+ d^2*h^2*z^4*#1^3 & , 3]^2 + 2*d*h*z* Root[c^2*z - 2*c*d*h*#1 - 4*c*d*h*z*#1 +
c^2*z^2*#1 - 2*c*d*h*z^2*#1 + 2*c*d*h* #1^2 - c^2*z*#1^2 + 2*c*d*h*z*#1^2 -
2*c*d*h* z^2*#1^2 - 2*c*d*h*z^3* #1^2 + d^2*h^2*#1^3 + 4*d^2*h^2*z*#1^3 +
6*d^2*h^2*z^2*#1^3 + 4*d^2*h^2*z^3*#1^3 + d^2*h^2*z^4*#1^3 & , 3]^2 + d*h*z^2*
Root[c^2*z - 2*c*d*h*#1 - 4*c*d*h*z*#1 + c^2*z^2*#1 - 2*c*d*h*z^2*#1 + 2*c*d*h*
#1^2 - c^2*z*#1^2 + 2*c*d*h*z*#1^2 - 2*c*d*h* z^2*#1^2 - 2*c*d*h*z^3* #1^2 +
d^2*h^2*#1^3 + 4*d^2*h^2*z*#1^3 + 6*d^2*h^2*z^2*#1^3 + 4*d^2*h^2*z^3*#1^3 +
d^2*h^2*z^4*#1^3 & , 3]^2 + c*Sqrt[1 + z*Root[ c^2*z - 2*c*d*h*#1 - 4*c*d*h*z*#1
+ c^2*z^2* #1 - 2*c*d*h*z^2*#1 + 2*c*d*h*#1^2 - c^2*z* #1^2 + 2*c*d*h*z*#1^2 -
2*c*d*h*z^2*#1^2 - 2*c*d*h*z^3*#1^2 + d^2*h^2*#1^3 + 4*d^2*h^2* z*#1^3 +
6*d^2*h^2*z^2* #1^3 + 4*d^2*h^2*z^3* #1^3 + d^2*h^2*z^4* #1^3 & , 3]] -
c*Root[c^2*z - 2*c*d*h*#1 - 4*c*d*h*z*#1 + c^2*z^2*#1 - 2*c*d*h*z^2*#1 +
2*c*d*h* #1^2 - c^2*z*#1^2 + 2*c*d*h*z*#1^2 - 2*c*d*h*z^2* #1^2 -
2*c*d*h*z^3*#1^2 + d^2*h^2*#1^3 + 4*d^2*h^2*z* #1^3 + 6*d^2*h^2*z^2*#1^3 +
4*d^2*h^2*z^3*#1^3 + d^2*h^2*z^4*#1^3 & , 3]* Sqrt[1 + z*Root[ c^2*z -
2*c*d*h*#1 - 4*c*d*h*z*#1 + c^2*z^2* #1 - 2*c*d*h*z^2*#1 + 2*c*d*h*#1^2 - c^2*z*
#1^2 + 2*c*d*h*z*#1^2 - 2*c*d*h*z^2*#1^2 - 2*c*d*h*z^3*#1^2 + d^2*h^2*#1^3 +
4*d^2*h^2* z*#1^3 + 6*d^2*h^2*z^2* #1^3 + 4*d^2*h^2*z^3* #1^3 + d^2*h^2*z^4*
#1^3 & , 3]])/ (c*(-1 + Root[c^2*z - 2*c*d*h* #1 - 4*c*d*h*z*#1 + c^2*z^2*#1 -
2*c*d*h*z^2* #1 + 2*c*d*h*#1^2 - c^2*z*#1^2 + 2*c*d*h*z* #1^2 - 2*c*d*h*z^2*#1^2
- 2*c*d*h*z^3*#1^2 + d^2*h^2*#1^3 + 4*d^2*h^2*z* #1^3 + 6*d^2*h^2*z^2*#1^3 +
4*d^2*h^2*z^3*#1^3 + d^2*h^2*z^4*#1^3 & , 3])) && q == Root[c^2*z - 2*c*d*h*#1 -
4*c*d*h*z*#1 + c^2*z^2*#1 - 2*c*d*h*z^2*#1 + 2*c*d*h*#1^2 - c^2*z*#1^2 +
2*c*d*h*z*#1^2 - 2*c*d*h*z^2*#1^2 - 2*c*d*h*z^3* #1^2 + d^2*h^2*#1^3 +
4*d^2*h^2*z*#1^3 + 6*d^2*h^2* z^2*#1^3 + 4*d^2*h^2*z^3* #1^3 + d^2*h^2*z^4*#1^3
& , 3]) || (c != 0 && h != 0 && z == 0 && d == 0 && q == 1) || (c*z != 0 && -2 +
z - Sqrt[4 + z^2] != 0 && 0 == (4 - 2*z + 2*z^2 + 2*Sqrt[4 + z^2] - 2*z*Sqrt[4 +
z^2] - 2*Sqrt[2]*Sqrt[2 + z^2 - z*Sqrt[4 + z^2]] + Sqrt[2]*z*Sqrt[2 + z^2 -
z*Sqrt[4 + z^2]] - Sqrt[2]*Sqrt[4 + z^2]* Sqrt[2 + z^2 - z*Sqrt[4 + z^2]])/
(2*(2 - z + Sqrt[4 + z^2])) && 1 + z != 0 && h == 0 && q == (z - Sqrt[4 +
z^2])/2) || (c*z != 0 && -2 + z - Sqrt[4 + z^2] != 0 && 0 == (4 - 2*z + 2*z^2 +
2*Sqrt[4 + z^2] - 2*z*Sqrt[4 + z^2] - 2*Sqrt[2]*Sqrt[2 + z^2 - z*Sqrt[4 + z^2]]
+ Sqrt[2]*z*Sqrt[2 + z^2 - z*Sqrt[4 + z^2]] - Sqrt[2]*Sqrt[4 + z^2]* Sqrt[2 +
z^2 - z*Sqrt[4 + z^2]])/ (2*(2 - z + Sqrt[4 + z^2])) && h + h*z != 0 && d == 0
&& q == (z - Sqrt[4 + z^2])/2) || (c*z != 0 && -2 + z + Sqrt[4 + z^2] != 0 && 0
== (-4 + 2*z - 2*z^2 + 2*Sqrt[4 + z^2] - 2*z*Sqrt[4 + z^2] + 2*Sqrt[2]*Sqrt[2 +
z^2 + z*Sqrt[4 + z^2]] - Sqrt[2]*z*Sqrt[2 + z^2 + z*Sqrt[4 + z^2]] -
Sqrt[2]*Sqrt[4 + z^2]* Sqrt[2 + z^2 + z*Sqrt[4 + z^2]])/ (2*(-2 + z + Sqrt[4 +
z^2])) && 1 + z != 0 && h == 0 && q == (z + Sqrt[4 + z^2])/2) || (c*z != 0 && -2
+ z + Sqrt[4 + z^2] != 0 && 0 == (-4 + 2*z - 2*z^2 + 2*Sqrt[4 + z^2] -
2*z*Sqrt[4 + z^2] + 2*Sqrt[2]*Sqrt[2 + z^2 + z*Sqrt[4 + z^2]] - Sqrt[2]*z*Sqrt[2
+ z^2 + z*Sqrt[4 + z^2]] - Sqrt[2]*Sqrt[4 + z^2]* Sqrt[2 + z^2 + z*Sqrt[4 +
z^2]])/ (2*(-2 + z + Sqrt[4 + z^2])) && h + h*z != 0 && d == 0 && q == (z +
Sqrt[4 + z^2])/2)

So, for example, if you look at the first line of the "solution" you see
(z == -1 && c == 0) || lots_of_other_stuff.

Thus if we substitute -1 for z and 0 for c into your original equation we get
c (q*z + (-1 + q) (-1 + Sqrt[1 + q*z])) == d*(h*q^2 (1 + z)^2)
0 (q*z + (-1 + q) (-1 + Sqrt[1 + q* -1])) == d*(h*q^2 (1 + -1)^2)
0 (q*z + (-1 + q) (-1 + Sqrt[1 + q* -1])) == d*(h*q^2 0^2)
0 == 0

And likewise for all the other alternatives, some of which are very complicated.

Buried inside that are some
Root[expression_containing_#,n]
That is a much simpler way of saying that this is the nth root of the expression.
For your problem it appears possible to expand all those Root[] out to the full result containing radicals, but expands the size of this by several fold.

If you have any additional information, like whether some of your variables are or are not equal to zero, whether they are positive, whether they are or are not equal to each other, or that 1+q*z will be greater than or equal to zero so that the square root will not be complex, etc., then it might be possible to substantially simplify that. But there are no guarantees.

Unless you can provide additional information or you can identify one of the alternatives that captures your problem, both those probably being close to the same thing, it looks like such a "solution" probably isn't going to be of any use to you.

Simplify[] can sometimes reduce the size of things like this by perhaps half, but sometimes that will lose a solution.
 
Last edited:

Related to Is Mathematica able to solve expressions such as

1. What is Mathematica and how does it solve expressions?

Mathematica is a computational software program used for mathematical, scientific, and technical computations. It uses advanced algorithms and symbolic computation to solve expressions and equations in a wide range of mathematical areas.

2. Can Mathematica solve any type of expression?

Mathematica can solve a wide variety of mathematical expressions, including algebraic, differential, integral, and statistical equations. It also has built-in functions for handling complex numbers, matrices, and other mathematical objects.

3. How accurate are the solutions provided by Mathematica?

The solutions provided by Mathematica are typically very accurate, as it uses high-precision numerical methods and symbolic computation to ensure precision and correctness. However, the accuracy may vary depending on the complexity of the expression and the input data.

4. Is Mathematica able to handle large expressions or datasets?

Yes, Mathematica is designed to handle large expressions and datasets. It has efficient memory management and can handle complex computations on large datasets without compromising speed or accuracy.

5. Can Mathematica be used for data analysis and visualization?

Yes, Mathematica has built-in functions for data analysis and visualization. It can import data from various sources, perform statistical analysis, and create interactive visualizations. It is widely used in scientific research and data analysis in various fields.

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