# Find the sum of real roots

#### anemone

##### MHB POTW Director
Staff member
Determine the sum of real roots of the equation $q^4-7q^3+14q^2-14q+4=0$.

#### mathbalarka

##### Well-known member
MHB Math Helper
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

##### MHB POTW Director
Staff member
Awesome, Balarka and thanks for participating!

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

#### mathbalarka

##### Well-known member
MHB Math Helper
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

##### Well-known member
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

##### Well-known member
MHB Math Helper
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.

#### anemone

##### MHB POTW Director
Staff member
$x4−4x2+4−7x(x2+2)+10x2=(x2+1)2−7x(x2+2)+10x2$

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!

#### MarkFL

Staff member
$x4−4x2+4−7x(x2+2)+10x2=(x2+1)2−7x(x2+2)+10x2$

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
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 $x^{2y}$