Finding general solution of Radical Equation

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SUMMARY

The discussion centers on the feasibility of finding a general solution for radical equations, particularly those of the form A1 + A2x + A3√(B1 + B2x + B3x^(3/2) + B4√x + B5x^2) + A4√(C1 + C2x + C3x^(3/2) + C4√x + C5x^2) = 0. Participants conclude that while isolating square roots and squaring both sides can simplify the equation, a general solution does not exist for polynomial equations of degree 5 or higher, as per the Abel-Ruffini theorem. The complexity increases with the introduction of higher degree terms, suggesting that solutions may not be expressible in terms of radicals.

PREREQUISITES
  • Understanding of radical equations and their properties.
  • Familiarity with polynomial equations and their degrees.
  • Knowledge of the Abel-Ruffini theorem regarding polynomial solvability.
  • Basic algebraic manipulation skills, including isolating terms and squaring equations.
NEXT STEPS
  • Study the Abel-Ruffini theorem in detail to understand its implications on polynomial equations.
  • Learn techniques for solving radical equations, including isolation and squaring methods.
  • Explore numerical methods for approximating solutions to complex radical equations.
  • Investigate alternative forms of solutions for higher-degree equations, such as using numerical solvers or graphical methods.
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Mathematicians, algebra students, and anyone interested in the complexities of solving radical equations and polynomial equations of higher degrees.

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Before trying to find out the general solution of a radical equation; I would first like to know if it can be found?
For example I have a equation of the form
<br /> \text{A1}+\text{A2} x + \text{A3}\sqrt{\text{B1}+\text{B2} x+\text{B3} x^{\frac{3}{2}}+\text{B4}\sqrt{x}+\text{B5} x^2}+ \text{A4}\sqrt{\text{C1}+\text{C2} x+\text{C3} x^{\frac{3}{2}}+\text{C4}\sqrt{x}+\text{C5} x^2}=0
Can I find x in terms of the Constants A1,A2 etc?
What is the general view on deciding whether a general solution to radical equation exist or not?
I tried searching, but couldn't find out the answer regarding radical equation.
For polynomial equation though, I learned that a general solution doesn't exist for polynomials of degree 5 or higher.
http://en.wikipedia.org/wiki/Abel–Ruffini_theorem
 
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1. Isolate one of the square roots.
2. Square both sides which will leave a single square root.
3. Isolate that square root.
4. Now you will have an equation involving powers of x of 2 3/2, 1, and 1/2. Let y= x1/2 so that you have a polynomial involving y4, y3, y2, and y.

(There cannot be a "general solution for a polynomial equation" of degee greater than 5 involving only powers and roots of the coefficients because they may have solution that cannot be written in terms of roots.)
 
HallsofIvy said:
1. Isolate one of the square roots.
2. Square both sides which will leave a single square root.
3. Isolate that square root and Square both sides
4. Now you will have an equation involving powers of x of 2 3/2, 1, and 1/2. Let y= x1/2 so that you have a polynomial involving y4, y3, y2, and y.

(There cannot be a "general solution for a polynomial equation" of degee greater than 5 involving only powers and roots of the coefficients because they may have solution that cannot be written in terms of roots.)
(I made a little amendments). Thanks.
1.\text{A1}+\text{A2} x + \text{A3}\sqrt{\text{B1}+\text{B2} x+\text{B3} x^{\frac{3}{2}}+\text{B4}\sqrt{x}+\text{B5} x^2}= \text{A4}\sqrt{\text{C1}+\text{C2} x+\text{C3} x^{\frac{3}{2}}+\text{C4}\sqrt{x}+\text{C5} x^2}
2.\left(\text{A1}+\text{A2} x + \text{A3}\sqrt{\text{B1}+\text{B2} x+\text{B3} x^{\frac{3}{2}}+\text{B4}\sqrt{x}+\text{B5} x^2}\right)^2= \left(\text{A4}\sqrt{\text{C1}+\text{C2} x+\text{C3} x^{\frac{3}{2}}+\text{C4}\sqrt{x}+\text{C5} x^2}\right)^2
\left(\text{A1}^2+\text{A3}^2 \text{B1}+\text{A3}^2 \text{B4} \sqrt{x}+2 \text{A1} \text{A2} x+\text{A3}^2 \text{B2} x+\text{A3}^2 \text{B3} x^{3/2}+\text{A2}^2 x^2+\text{A3}^2 \text{B5} x^2+\text{A3} (2 \text{A1}+2 \text{A2} x) \sqrt{\text{B1}+\text{B4} \sqrt{x}+\text{B2} x+\text{B3} x^{3/2}+\text{B5} x^2}\right)=\text{A4}^2 \text{C1}+\text{A4}^2 \text{C4} \sqrt{x}+\text{A4}^2 \text{C2} x+\text{A4}^2 \text{C3} x^{3/2}+\text{A4}^2 \text{C5} x^2
3.\text{A3} (2 \text{A1}+2 \text{A2} x) \sqrt{\text{B1}+\text{B4} \sqrt{x}+\text{B2} x+\text{B3} x^{3/2}+\text{B5} x^2}=\left(\text{A4}^2 \text{C1}+\text{A4}^2 \text{C4} \sqrt{x}+\text{A4}^2 \text{C2} x+\text{A4}^2 \text{C3} x^{3/2}+\text{A4}^2 \text{C5} x^2\right)-\left(\text{A1}^2+\text{A3}^2 \text{B1}+\text{A3}^2 \text{B4} \sqrt{x}+2 \text{A1} \text{A2} x+\text{A3}^2 \text{B2} x+\text{A3}^2 \text{B3} x^{3/2}+\text{A2}^2 x^2+\text{A3}^2 \text{B5} x^2\right)
\left(\text{A3} (2 \text{A1}+2 \text{A2} x) \sqrt{\text{B1}+\text{B4} \sqrt{x}+\text{B2} x+\text{B3} x^{3/2}+\text{B5} x^2}\right)^2=\left(\left(\text{A4}^2 \text{C1}+\text{A4}^2 \text{C4} \sqrt{x}+\text{A4}^2 \text{C2} x+\text{A4}^2 \text{C3} x^{3/2}+\text{A4}^2 \text{C5} x^2\right)-\left(\text{A1}^2+\text{A3}^2 \text{B1}+\text{A3}^2 \text{B4} \sqrt{x}+2 \text{A1} \text{A2} x+\text{A3}^2 \text{B2} x+\text{A3}^2 \text{B3} x^{3/2}+\text{A2}^2 x^2+\text{A3}^2 \text{B5} x^2\right)\right)^2
4.4 \text{A1}^2 \text{A3}^2 \text{B1}+4 \text{A1}^2 \text{A3}^2 \text{B4} \sqrt{x}+8 \text{A1} \text{A2} \text{A3}^2 \text{B1} x+4 \text{A1}^2 \text{A3}^2 \text{B2} x+4 \text{A1}^2 \text{A3}^2 \text{B3} x^{3/2}+8 \text{A1} \text{A2} \text{A3}^2 \text{B4} x^{3/2}+4 \text{A2}^2 \text{A3}^2 \text{B1} x^2+8 \text{A1} \text{A2} \text{A3}^2 \text{B2} x^2+4 \text{A1}^2 \text{A3}^2 \text{B5} x^2+8 \text{A1} \text{A2} \text{A3}^2 \text{B3} x^{5/2}+4 \text{A2}^2 \text{A3}^2 \text{B4} x^{5/2}+4 \text{A2}^2 \text{A3}^2 \text{B2} x^3+8 \text{A1} \text{A2} \text{A3}^2 \text{B5} x^3+4 \text{A2}^2 \text{A3}^2 \text{B3} x^{7/2}+4 \text{A2}^2 \text{A3}^2 \text{B5} x^4=\text{A1}^4+2 \text{A1}^2 \text{A3}^2 \text{B1}+\text{A3}^4 \text{B1}^2-2 \text{A1}^2 \text{A4}^2 \text{C1}-2 \text{A3}^2 \text{A4}^2 \text{B1} \text{C1}+\text{A4}^4 \text{C1}^2+2 \text{A1}^2 \text{A3}^2 \text{B4} \sqrt{x}+2 \text{A3}^4 \text{B1} \text{B4} \sqrt{x}-2 \text{A3}^2 \text{A4}^2 \text{B4} \text{C1} \sqrt{x}-2 \text{A1}^2 \text{A4}^2 \text{C4} \sqrt{x}-2 \text{A3}^2 \text{A4}^2 \text{B1} \text{C4} \sqrt{x}+2 \text{A4}^4 \text{C1} \text{C4} \sqrt{x}+4 \text{A1}^3 \text{A2} x+4 \text{A1} \text{A2} \text{A3}^2 \text{B1} x+2 \text{A1}^2 \text{A3}^2 \text{B2} x+2 \text{A3}^4 \text{B1} \text{B2} x+\text{A3}^4 \text{B4}^2 x-4 \text{A1} \text{A2} \text{A4}^2 \text{C1} x-2 \text{A3}^2 \text{A4}^2 \text{B2} \text{C1} x-2 \text{A1}^2 \text{A4}^2 \text{C2} x-2 \text{A3}^2 \text{A4}^2 \text{B1} \text{C2} x+2 \text{A4}^4 \text{C1} \text{C2} x-2 \text{A3}^2 \text{A4}^2 \text{B4} \text{C4} x+\text{A4}^4 \text{C4}^2 x+2 \text{A1}^2 \text{A3}^2 \text{B3} x^{3/2}+2 \text{A3}^4 \text{B1} \text{B3} x^{3/2}+4 \text{A1} \text{A2} \text{A3}^2 \text{B4} x^{3/2}+2 \text{A3}^4 \text{B2} \text{B4} x^{3/2}-2 \text{A3}^2 \text{A4}^2 \text{B3} \text{C1} x^{3/2}-2 \text{A3}^2 \text{A4}^2 \text{B4} \text{C2} x^{3/2}-2 \text{A1}^2 \text{A4}^2 \text{C3} x^{3/2}-2 \text{A3}^2 \text{A4}^2 \text{B1} \text{C3} x^{3/2}+2 \text{A4}^4 \text{C1} \text{C3} x^{3/2}-4 \text{A1} \text{A2} \text{A4}^2 \text{C4} x^{3/2}-2 \text{A3}^2 \text{A4}^2 \text{B2} \text{C4} x^{3/2}+2 \text{A4}^4 \text{C2} \text{C4} x^{3/2}+6 \text{A1}^2 \text{A2}^2 x^2+2 \text{A2}^2 \text{A3}^2 \text{B1} x^2+4 \text{A1} \text{A2} \text{A3}^2 \text{B2} x^2+\text{A3}^4 \text{B2}^2 x^2+2 \text{A3}^4 \text{B3} \text{B4} x^2+2 \text{A1}^2 \text{A3}^2 \text{B5} x^2+2 \text{A3}^4 \text{B1} \text{B5} x^2-2 \text{A2}^2 \text{A4}^2 \text{C1} x^2-2 \text{A3}^2 \text{A4}^2 \text{B5} \text{C1} x^2-4 \text{A1} \text{A2} \text{A4}^2 \text{C2} x^2-2 \text{A3}^2 \text{A4}^2 \text{B2} \text{C2} x^2+\text{A4}^4 \text{C2}^2 x^2-2 \text{A3}^2 \text{A4}^2 \text{B4} \text{C3} x^2-2 \text{A3}^2 \text{A4}^2 \text{B3} \text{C4} x^2+2 \text{A4}^4 \text{C3} \text{C4} x^2-2 \text{A1}^2 \text{A4}^2 \text{C5} x^2-2 \text{A3}^2 \text{A4}^2 \text{B1} \text{C5} x^2+2 \text{A4}^4 \text{C1} \text{C5} x^2+4 \text{A1} \text{A2} \text{A3}^2 \text{B3} x^{5/2}+2 \text{A3}^4 \text{B2} \text{B3} x^{5/2}+2 \text{A2}^2 \text{A3}^2 \text{B4} x^{5/2}+2 \text{A3}^4 \text{B4} \text{B5} x^{5/2}-2 \text{A3}^2 \text{A4}^2 \text{B3} \text{C2} x^{5/2}-4 \text{A1} \text{A2} \text{A4}^2 \text{C3} x^{5/2}-2 \text{A3}^2 \text{A4}^2 \text{B2} \text{C3} x^{5/2}+2 \text{A4}^4 \text{C2} \text{C3} x^{5/2}-2 \text{A2}^2 \text{A4}^2 \text{C4} x^{5/2}-2 \text{A3}^2 \text{A4}^2 \text{B5} \text{C4} x^{5/2}-2 \text{A3}^2 \text{A4}^2 \text{B4} \text{C5} x^{5/2}+2 \text{A4}^4 \text{C4} \text{C5} x^{5/2}+4 \text{A1} \text{A2}^3 x^3+2 \text{A2}^2 \text{A3}^2 \text{B2} x^3+\text{A3}^4 \text{B3}^2 x^3+4 \text{A1} \text{A2} \text{A3}^2 \text{B5} x^3+2 \text{A3}^4 \text{B2} \text{B5} x^3-2 \text{A2}^2 \text{A4}^2 \text{C2} x^3-2 \text{A3}^2 \text{A4}^2 \text{B5} \text{C2} x^3-2 \text{A3}^2 \text{A4}^2 \text{B3} \text{C3} x^3+\text{A4}^4 \text{C3}^2 x^3-4 \text{A1} \text{A2} \text{A4}^2 \text{C5} x^3-2 \text{A3}^2 \text{A4}^2 \text{B2} \text{C5} x^3+2 \text{A4}^4 \text{C2} \text{C5} x^3+2 \text{A2}^2 \text{A3}^2 \text{B3} x^{7/2}+2 \text{A3}^4 \text{B3} \text{B5} x^{7/2}-2 \text{A2}^2 \text{A4}^2 \text{C3} x^{7/2}-2 \text{A3}^2 \text{A4}^2 \text{B5} \text{C3} x^{7/2}-2 \text{A3}^2 \text{A4}^2 \text{B3} \text{C5} x^{7/2}+2 \text{A4}^4 \text{C3} \text{C5} x^{7/2}+\text{A2}^4 x^4+2 \text{A2}^2 \text{A3}^2 \text{B5} x^4+\text{A3}^4 \text{B5}^2 x^4-2 \text{A2}^2 \text{A4}^2 \text{C5} x^4-2 \text{A3}^2 \text{A4}^2 \text{B5} \text{C5} x^4+\text{A4}^4 \text{C5}^2 x^4

But now there are x^4,x^{\frac{7}{2}},x^3,x^{\frac{5}{2}},x^2,x^{\frac{3}{2}},x,\sqrt{x}
If I replace x = y^2 then I will have polynomial of degree 8.
So it appears the equation won't have general solution. (Atleast not in terms of roots and powers as you said). But I wonder in what form I might get the solution, if at all.
 

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