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- Feb 14, 2012

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

- Thread starter anemone
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- Feb 14, 2012

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

- 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

.

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

.

Last edited:

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- #3

- Feb 14, 2012

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

- Mar 22, 2013

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Thanks, and I'll participate in almost everything you have about theory of polynomial equations and number theoryanemone said:Awesome, Balarka and thanks for participating!

- Mar 31, 2013

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Here is my solution :

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

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

.

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

- Mar 22, 2013

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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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- Feb 14, 2012

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Hi

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

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Hello

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

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 $x^{2y}$