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Find the sum of real roots

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anemone

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Feb 14, 2012
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Determine the sum of real roots of the equation $q^4-7q^3+14q^2-14q+4=0$.
 

mathbalarka

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Mar 22, 2013
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Here is my solution :

\(\displaystyle \begin{aligned} x^4 - 7x^3 + 14x^2 - 14x + 4 &= x^4 - 7x(x^2 + 2) + 14x^2 + 4 \\ &= x^4 - 4x^2 + 4 - 7x(x^2 + 2) + 10x^2 \\ &= (x^2 + 2)^2 - 7x(x^2 + 2) + 10x^2 \end{aligned}\)

Now, set $y = x^2 + 2$, so that the polynomial transforms into

$y^2 - 7xy + 10x^2 = (y - 5x)(y - 2x) = (x^2 - 5x + 2)(x^2 - 2x + 2)$

The discriminant of the former factor is 17 and the latter is -4, implying the former factor contains all the real roots of the quartic. Hence, the sum of all real roots are 5.


Balarka
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anemone

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Feb 14, 2012
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Awesome, Balarka and thanks for participating!

I really like your method because my approach is a bit convoluted.:eek:
 

mathbalarka

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Mar 22, 2013
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anemone said:
Awesome, Balarka and thanks for participating!
Thanks, and I'll participate in almost everything you have about theory of polynomial equations and number theory ;)
 

kaliprasad

Well-known member
Mar 31, 2013
1,322
Here is my solution :

\(\displaystyle \begin{aligned} x^4 - 7x^3 + 14x^2 - 14x + 4 &= x^4 - 7x(x^2 + 2) + 14x^2 + 4 \\ &= x^4 - 4x^2 + 4 - 7x(x^2 + 2) + 10x^2 \\ &= (x^2 + 1)^2 - 7x(x^2 + 2) + 10x^2 \end{aligned}\)

Now, set $y = x^2 + 1$, so that the polynomial transforms into

$y^2 - 7xy + 10x^2 = (y - 5x)(y - 2x) = (x^2 - 5x + 2)(x^2 - 2x + 2)$

The discriminant of the former factor is 17 and the latter is -4, implying the former factor contains all the real roots of the quartic. Hence, the sum of all real roots are 5.


Balarka
.
$x4−4x2+4−7x(x2+2)+10x2=(x2+1)2−7x(x2+2)+10x2$

Now, set $y=x2+1$, so that the polynomial transforms into

the above should be
$x4−4x2+4−7x(x2+2)+10x2=(x2+2)2−7x(x2+2)+10x2$

Now, set y=x2+2, so that the polynomial transforms into
 

mathbalarka

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MHB Math Helper
Mar 22, 2013
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Yes, thanks for notifying me the typo.

PS : Were working on something very complicated when I posted it, a proof of primality for a class of 'sum of cubes equal to square of sum' multisets ... apologies, anyway.
 
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anemone

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Feb 14, 2012
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$x4−4x2+4−7x(x2+2)+10x2=(x2+1)2−7x(x2+2)+10x2$

Now, set $y=x2+1$, so that the polynomial transforms into

the above should be
$x4−4x2+4−7x(x2+2)+10x2=(x2+2)2−7x(x2+2)+10x2$

Now, set y=x2+2, so that the polynomial transforms into
Hi kaliprasad, thanks for being a fresh pair of eyes for me! :eek:
 

MarkFL

Administrator
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Feb 24, 2012
13,775
$x4−4x2+4−7x(x2+2)+10x2=(x2+1)2−7x(x2+2)+10x2$

Now, set $y=x2+1$, so that the polynomial transforms into

the above should be
$x4−4x2+4−7x(x2+2)+10x2=(x2+2)2−7x(x2+2)+10x2$

Now, set y=x2+2, so that the polynomial transforms into
Hello kaliprasad,

To write exponents in $\LaTeX$, use the caret character "^". If an exponent has more than 1 character, then enclose it in curly braces. For example:

x^2 gives $x^2$

x^{2y} gives $x^{2y}$