Can You Solve the Square Root Equation with Variables?

[liskawc]

[SOLVED] how to solve and equation?

Homework Statement

sqrt(ax^2 +bx) - sqrt(cx^2 + dx) = e
where you know a,b,c,d and e

Homework Equations

i don't know

The Attempt at a Solution

i put the entire equation on the power of 2 and then simplified adn then repeated again ... and i got an equation of 4th grade (ax^4+bx^3+...=0)
and well i can't solve it so i retuned back to my original problem :)

 

[Astronuc]

If the first sqrt argument was a form (x+p)² and the second sqrt argument was (x+q)², then one would have a form

x+p + x+q = e

or x = 1/2 (e - p - q) and don't forget roots +/-

 

[liskawc]

hmm sadly its not a full square :(

solved it with taylor series (got and estimate and its better then nothing)

 

[kbaumen]

I'm talking only about the case when a, b, c, d and e are integers.

First of all, substract e from both sides you get the form (let d - e = f) ax$$^{4}$$ + bx$$^{3}$$ + cx$$^{2}$$ + f = 0

Then you have to factor the equation to a(x - x$$_{1}$$)(x - x$$_{2}$$)(x - x$$_{3}$$)(x - x$$_{4}$$) = 0.

To do this you must first figure out one of the roots. All roots are in the set of $$\frac{d}{a}$$ divisors, so what I usually do is try to input all those divisors as x. Now, when you find one root, you must input it in Horner's scheme.

1)In Horner's scheme you put in the line A all coefficients of x$$^{n}$$ - a, b, c and d. In position C1 (look at the example) you place the root you guessed. Now rewrite the coefficient from A2 to C2. Multiply C2 with the root (B1) and write the number at B3. Now add to it A3 and write it at C3. Multiply C3 with the root and rewrite it at B4. Add A4 to B4 and rewrite the sum at C4. Now multiply it with the root and rewrite the product at B5. etc. Do you see the algorithm? If no, then http://en.wikipedia.org/wiki/Horner_scheme" it might me explained a bit better. Anyway, if C6 is not 0 then check your calculations again or the root might be not a root. Now the C line are coefficients for (x - x$$_{1}$$)(C2x$$^{3}$$ + C3x$$^{2}$$ + C4x + C5) = 0. Now factor the part in second brackets until you reach the form previously mentioned where x$$_{1}$$, x$$_{2}$$ etc. are the roots you need.

Here's an example:

5x$$^{4$$ - 3x$$^{3}$$ - 4x$$^{2}$$ - 3x + 5 = 0

I find that one of the roots is x$$_{1}$$ = 1 so

1. 2. 3. 4. 5. 6. ==== position numbers
---------------------------------
| 5 -3 -4 -3 5 (A)
|
1 | 5 2 -2 -5 (B)
|-----------------------------
5 2 -2 -5 0 (C)

Now I have 5(x - 1)(5x$$^{3}$$ + 2x$$^{2}$$ - 2x - 5) = 0

On 5x$$^{3}$$ + 2x$$^{2}$$ - 2x - 5 I use Horner's scheme again. I guess that one root x$$_{2}$$ = 1 and...

| 5 2 -2 -5
|
1 | 5 7 5
|--------------------------
| 5 7 5 0

Now I have 25(x - 1)(x - 1)(5x$$^{2}$$ + 7x + 5) = 0

You probably now how to factor 5x$$^{2}$$ + 7x + 5 so I get

125(x - 1)(x - 1)(x - (7 + 2$$\sqrt{6}$$)/2)(x - (7 - 2$$\sqrt{6}$$)/2) = 0

If the result is 0 then at least one of the multipliers must be 0. So x$$_{1}$$ = x$$_{2}$$ = 1; x$$_{3}$$ = (7 + 2$$\sqrt{6}$$)/2; x$$_{4}$$ = (7 - 2$$\sqrt{6}$$)/2).

I hope that helps.

Whew, that tok a lot of time. If something is unclear, and it probably will be from what I've written, feel free to ask.

 

[liskawc]

tried it and it works :D thank you (does it work for x^n+... aswell?)

 

[kbaumen]

tried it and it works :D thank you (does it work for x^n+... aswell?)

Yes, in x$$^{n}$$+... n is the max number of roots by the way.

 

[liskawc]

cool :)

thanks