MHB Equations with rational degrees

[mathmaniac1]

Is there a way to make these a polynomial?
1)$$\sqrt{A}+\sqrt{B}+\sqrt{C}=D$$
2)A^1/3 + B^1/3 = C
Or how else can it be solved?

 

[MarkFL]

One can always appropriately arrange, then raise both sides to an exponent that changes exponents to integral values. This may have to be repeated to complete the process. If course, then solutions must be checked, since extraneous solutions may be introduced.

 

[mathmaniac1]

Edit

1)$$\sqrt{A}+\sqrt{B}+\sqrt{C}+\sqrt{D}=E$$

Please show how to solve modified 1) and 2),i.e, make them polynomial

 

[MarkFL]

I'll demonstrate the process for 2), and this will give you an idea how to do 1).

2.) $$\sqrt[3]{A}+\sqrt[3]{B}=C$$

Cube both sides to get:

$$A+3\sqrt[3]{A^2B}+3\sqrt[3]{AB^2}+B=C^3$$

$$3\sqrt[3]{AB}\left(\sqrt[3]{A}+\sqrt[3]{B} \right)=C^3-(A+B)$$

The original equation allows us to simplify as:

$$3C\sqrt[3]{AB}=C^3-(A+B)$$

Cube again:

$$27ABC^3=C^9-3C^6(A+B)+3C^3(A+B)^2+(A+B)^3$$

 

[mathmaniac1]

Nice
Please give an answer for 1) too
I tried and failed

General Question:Can this be done for any eq with rational powers?

 

[HallsofIvy]

Nice
Please give an answer for 1) too
I tried and failed

General Question:Can this be done for any eq with rational powers?

It should be obvious that the answer is yes.

As for your previous question please show what you tried.

 

[mathmaniac1]

Squaring,we get
$$2\sqrt{AB}+2\sqrt{AC}+2\sqrt{AD}+2\sqrt{BC}+2\sqrt{BD}+2\sqrt{CD}...$$

I can take common $$2\sqrt{A}$$and sub for $$\sqrt{B}+\sqrt{C}+\sqrt{D}$$

Whatever I tried to do then didn't help...

Whoever's going to reply only needs to show how to reduce the number of rooted terms,make it less than 4...
I think there's some trick or maybe just my problem.

Please help

 

[Opalg]

Squaring,we get
$$2\sqrt{AB}+2\sqrt{AC}+2\sqrt{AD}+2\sqrt{BC}+2\sqrt{BD}+2\sqrt{CD}...$$

I can take common $$2\sqrt{A}$$and sub for $$\sqrt{B}+\sqrt{C}+\sqrt{D}$$

Whatever I tried to do then didn't help...

Whoever's going to reply only needs to show how to reduce the number of rooted terms,make it less than 4...
I think there's some trick or maybe just my problem.

Please help

Take one term across to the other side before squaring: $\sqrt A + \sqrt B = D - \sqrt C$.

 

[mathmaniac1]

Take one term across to the other side before squaring: $\sqrt A + \sqrt B = D - \sqrt C$.

I meant the modified 1
See my edit.

 

[Opalg]

I meant the modified 1
See my edit.

[I should have read the whole thread – I didn't see the modification.]

Try an inductive method. Start with the equation $\sqrt{A}+\sqrt{B}+\sqrt{C}=F$. That only has three roots, so you (presumably) know how to remove them to get a polynomial equation in powers of $A$, $B$, $C$ and $F$. Now substitute $F = E - \sqrt D$ into that equation. You will then have an equation in which the only radical is $\sqrt D$. So collect all the terms with $\sqrt D$ onto one side of that equation and square again.

 

[I like Serena]

Is there a way to make these a polynomial?
1)$$\sqrt{A}+\sqrt{B}+\sqrt{C}=D$$
2)A^1/3 + B^1/3 = C
Or how else can it be solved?

Alternatively, substitute $A=x^6, B=y^6, C=z^6$ with $x,y,z \ge 0$.
Then you get:
1) $$x^3+y^3+z^3=D$$
2) $$x^2+ y^2= z^6$$

 

[mathmaniac1]

Alternatively, substitute $A=x^6, B=y^6, C=z^6$ with $x,y,z \ge 0$.
Then you get:
1) $$x^3+y^3+z^3=D$$
2) $$x^2+ y^2= z^6$$

No,I was thinking about A,B,C...representing polynomial functions on a variable (say x) but RHS a constant.There you can't do these kind of techniques when your purpose is to solve for x.

But nice (funny?) answer,for my less informative post.

 

[mathmaniac1]

[I should have read the whole thread – I didn't see the modification.]

Try an inductive method. Start with the equation $\sqrt{A}+\sqrt{B}+\sqrt{C}=F$. That only has three roots, so you (presumably) know how to remove them to get a polynomial equation in powers of $A$, $B$, $C$ and $F$. Now substitute $F = E - \sqrt D$ into that equation. You will then have an equation in which the only radical is $\sqrt D$. So collect all the terms with $\sqrt D$ onto one side of that equation and square again.

Very very nice trick,impressive!

using this,i can make any eq with only degree as 1/2 a polynomial no matter,how many terms there are...

I'll be soon bringing another question about rational degrees(maybe) and I think the general question can also be solved by induction.Right?
Let me see.