# Thread: A small doubt

1. ## A small doubt

Is there any way to find sq.root of a fourth degree polynomial?
In this question, x^4 -2x^3 +3x^2 -2x +1 ,how can we find. Sq.root?

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2. ## Re: A small doubt

I would write:

$\displaystyle x^4-2x^3+3x^2-2x+1=(x^2+ax+b)^2=x^4+2ax^3+(a^2+2b)x^2+2abx+b^2$

Equating coefficients, we obtain the system:

$\displaystyle 2a=-2\implies a=-1$

$\displaystyle a^2+2b=3\implies b=1$

And so we have:

$\displaystyle x^4-2x^3+3x^2-2x+1=(x^2-x+1)^2$

What do you then conclude?

3. ## Re: A small doubt

Originally Posted by MarkFL
I would write:

$\displaystyle x^4-2x^3+3x^2-2x+1=(x^2+ax+b)^2=x^4+2ax^3+(a^2+2b)x^2+2abx+b^2$

Equating coefficients, we obtain the system:

$\displaystyle 2a=-2\implies a=-1$

$\displaystyle a^2+2b=3\implies b=1$

And so we have:

$\displaystyle x^4-2x^3+3x^2-2x+1=(x^2-x+1)^2$

What do you then conclude?
Thats great. So we should use the principle that square of n termed polynomial have 2n terms.And then equate..
Thanks again bro..

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4. ## Re: A small doubt

Originally Posted by MarkFL
I would write:

$\displaystyle x^4-2x^3+3x^2-2x+1=(x^2+ax+b)^2=x^4+2ax^3+(a^2+2b)x^2+2abx+b^2$

Although it's natural to do the above, and I would expect someone to set it up that way, it could also be done this way:

You could write:

$\displaystyle x^4 - 2x^3 + 3x^2 - 2x + 1 \ = \ (-x^2 + ax + b)^2 \ = \ x^4 - 2ax^3 + (a^2 - 2b)x^2 + 2abx + b^2$

Equating coefficients, we obtain the system:

$\displaystyle -2a \ = \ -2 \ \implies \ a \ = \ 1$

$\displaystyle a^2 - 2b \ = \ 3 \ \implies \ b \ = \ -1$

And so we have:

$\displaystyle x^4 - 2x^3 + 3x^2 - 2x + 1 \ = \ (-x^2 + x - 1)^2$

__________________________________________________ _______________________

sbjsavio, it is not true in general that the square of an n-termed polynomial (restricting ourselves to one variable for this discussion)
has 2n terms (when the like terms have been combined).

Look at these counterexamples:

$\displaystyle (x)^2 \ = \ x^2$

$\displaystyle (x + 1)^2 \ = \ x^2 + 2x + 1$

$\displaystyle (x^2 + 4x - 8)^2 \ = \ x^4 + 8x^3 - 64x + 64$

__________________________________________________ ________

Originally Posted by sbjsavio
Is there any way to find sq.root of a fourth degree polynomial?
In this question, x^4 -2x^3 +3x^2 -2x +1 ,how can we find. Sq.root?
Also, there is a square root algorithm for polynomials that someone gives as an example here:

https://math.stackexchange.com/quest...f-a-polynomial

5. ## Re: A small doubt

Originally Posted by greg1313
Although it's natural to do the above, and I would expect someone to set it up that way, it could also be done this way:

You could write:

$\displaystyle x^4 - 2x^3 + 3x^2 - 2x + 1 \ = \ (-x^2 + ax + b)^2 \ = \ x^4 - 2ax^3 + (a^2 - 2b)x^2 + 2abx + b^2$

Equating coefficients, we obtain the system:

$\displaystyle -2a \ = \ -2 \ \implies \ a \ = \ 1$

$\displaystyle a^2 - 2b \ = \ 3 \ \implies \ b \ = \ -1$

And so we have:

$\displaystyle x^4 - 2x^3 + 3x^2 - 2x + 1 \ = \ (-x^2 + x - 1)^2$

__________________________________________________ _______________________

sbjsavio, it is not true in general that the square of an n-termed polynomial (restricting ourselves to one variable for this discussion)
has 2n terms (when the like terms have been combined).

Look at these counterexamples:

$\displaystyle (x)^2 \ = \ x^2$

$\displaystyle (x + 1)^2 \ = \ x^2 + 2x + 1$

$\displaystyle (x^2 + 4x - 8)^2 \ = \ x^4 + 8x^3 - 64x + 64$

__________________________________________________ ________

Also, there is a square root algorithm for polynomials that someone gives as an example here:

https://math.stackexchange.com/quest...f-a-polynomial
Oh,thanks greg ,this is easier

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6. ## Re: A small doubt

Originally Posted by sbjsavio
x^4 -2x^3 +3x^2 -2x +1 , how can we find. Sq.root?
On a timed test, with about a minute left, you'd go this way:
a^2 = x^4 -2x^3 +3x^2 -2x +1
a = sqrt(x^4 -2x^3 +3x^2 -2x +1)

Well, perhaps a point or two

7. ## Re: A small doubt

Originally Posted by DenisB
On a timed test, with about a minute left, you'd go this way:
a^2 = x^4 -2x^3 +3x^2 -2x +1
a = sqrt(x^4 -2x^3 +3x^2 -2x +1)

Well, perhaps a point or two
Yup ..

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