Finding general solution of Radical Equation

In summary, a radical equation is an equation with a variable within a radical sign that can be solved by isolating the radical term and raising both sides to the same power. The general solution of a radical equation is the set of all possible values for the variable, often expressed in interval notation. To find the general solution, isolate the radical term, eliminate the radical, and solve for the variable. It is possible for a radical equation to have extraneous solutions, which must be checked for and discarded. There may also be restrictions on the values of the variable in a radical equation, such as the radicand being greater than or equal to zero in an equation with a square root.
  • #1
I_am_learning
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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
[itex]
\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[/itex]
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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  • #2
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.)
 
  • #3
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.[itex]\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}[/itex]
2.[itex]\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[/itex]
[itex]\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[/itex]
3.[itex]\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)[/itex]
[itex]\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[/itex]
4.[itex]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[/itex]

But now there are [itex]x^4,x^{\frac{7}{2}},x^3,x^{\frac{5}{2}},x^2,x^{\frac{3}{2}},x,\sqrt{x}[/itex]
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.
 

What is a radical equation?

A radical equation is an equation that contains a variable within a radical sign, such as square root or cube root. These equations can be solved by isolating the radical term and raising both sides of the equation to the same power to eliminate the radical.

What is the general solution of a radical equation?

The general solution of a radical equation is the set of all possible values for the variable that satisfy the equation. This solution may include multiple solutions, and it is often expressed in interval notation.

How do I find the general solution of a radical equation?

To find the general solution of a radical equation, first isolate the radical term by performing any necessary operations on both sides of the equation. Then, raise both sides of the equation to the same power to eliminate the radical. Finally, solve for the variable to obtain the general solution.

Can a radical equation have extraneous solutions?

Yes, a radical equation can have extraneous solutions. These are solutions that do not satisfy the original equation, often due to the presence of a squared term when solving for a square root. It is important to check for extraneous solutions and discard them when solving radical equations.

Are there any restrictions on the values of the variable in a radical equation?

Yes, there may be restrictions on the values of the variable in a radical equation. For example, if the equation contains a square root, the radicand (the number under the square root) must be greater than or equal to zero. It is important to check for any restrictions on the variable when solving radical equations.

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