Find the complex solutions

We note that by the Sophie-Germain identity, $x^4+4y^4=(x^2+2xy+2y^2)(x^2-2xy-2y^2)$ , so $x^4+(4)(12^4)=(x^2+(2)(12)x+(2)(12^2))(x^2-(2)(12)x+(2)(12^2))$ $=(x^2+24x+288)(x^2-24x+288)$ . Hence $\frac{x^4+(4)(12^4)}{x^2+24x+288}=x^2-24x+288$ .

Thus the equation becomes $x^2-24x+288=144x-288$ , so $x^2-168x+576$ . Since it is not a square polynomial as its discriminant, $168^2-4(1)(576)=25920 \neq 0$ , it has $2$ distinct roots. By Vieta's Formulas, their sum is $-\frac{-168}{1}=\boxed{168}$ .

Look at the numerator of the fraction in the left side. Using Sophie-Germain Identity we have ;

$=x^{4}+4.12^{4}$

$=x^{4}+4.12^{4}-4.12^{2}x^{2}+4.12^{2}x^{2}$

$=x^{4}+4.12^{2}x^{2}+4.12^{4}-4.12^{2}x^{2}$

$=(x^{2}+2.12^{2})^{2}-4.12^{2}x^{2}$

$=(x^{2}+2.12x+2.12^{2})(x^{2}-2.12x+2.12^{2})$

$=(x^{2}+24x+288)(x^{2}-24x+288)$

Since after the equation is distributed, we have the numerator and denumerator are equal $(x^{2}+24x+288)$ , so we can ommit it. Next, continue it with simple algebra;

$144x-288=x^{2}-24x+288$

$x^{2}-168x+576=0$

Since we find all the complex roots (including the real roots), so we get the answer is 168.

QED

We suspect that the numerator of the left-hand side is factorable. Indeed, we can "complete the square" on this expression as follows: $\begin{aligned} x^4 + 4 \cdot 12^4 &= (x^4 + 4 \cdot 12^2 \cdot x + 4 \cdot 12^4) - 4 \cdot 12^2 \cdot x \\ &= (x^2 + 2 \cdot 12^2)^2 - 4 \cdot 12^2 \cdot x^2 \\ &= (x^2+288)^2 - (24x)^2 \\ &= (x^2+24x+288)(x^2-24x+288), \end{aligned}$ where the last step follows from the difference-of-squares factorization. Therefore, the given equation becomes $\dfrac{(x^2+24x+288)(x^2-24x+288)}{x^2+24x+288} = 144x-288,$ or simply $x^2-24x+288 = 144x - 288.$ Rearranging this equation gives $x^2-168x + 576 = 0.$ By Vieta's Formulas, the sum of the two distinct roots of this equation is therefore $\boxed{168}.$

From the Sophie-Germain Identity ,

$\displaystyle x^4+4\cdot 12^4=(x^2+2\cdot 12^2-2\cdot x\cdot 12)(x^2+2\cdot 12^2+2\cdot x\cdot 12)=$ $\displaystyle (x^2-24x+288)(x^2+24x+288)$

Note that $x^2+24x+288$ has no roots so it is completely okay to cancel it. Hence, the given equation simplifies to,

$\displaystyle x^2-24x+288=144x-288 \Rightarrow x^2-168x+576=0$

The sum of roots of the above quadratic is $\fbox{168}$ and hence the answer.

PS: Brilliant trolls you. $12^4=20736$ . :D :D

Simply do long division of the fraction. This causes the expression to simplify to:

$x^{2}-24x+288=144x-28$

Then rewrite and apply Vieta's Formula to find the sum of roots - i.e

$x^{2}-168x+576=0 \therefore \frac{-b}{a}=168$ = Sum of roots.

Let $f(x) = x^{4} + 4⋅12^{4}$

$f(x) = (x^{2} - 24x + 288 )(x^{2} + 24x + 288)$

Hence, now the given equation becomes,

$x^{2} - 24x + 288 = 144x - 288$

$x^{2} - 168x + 576 = 0$ ...........(1)

according to the equation given we cannot assume the case that $x^{2} + 24x + 288 = 0$

Hence, the roots of the equation is given only by (1)

From (1), according to the Vieta's formula sum of the roots of this equation is 168

$\frac{(x^4+4\times12^4)}{(x^2+24x+288)}=144x-288$

$\frac{(x+12-12i)(x+12+12i)(x-12-12i)(x-12+12i)}{(x+12-12i)(x+12+12i)}=144x-288$

$(x-12-12i)(x-12+12i)=144x-288$

$x^2-24x+288=144x-288$

$x^2-168+576=0$

Sum of all complex solutions $=$ Sum of roots $= -(-168) = 168$

$x^4 + 4*12^4 = \left(x^2-kx+2*12^2\right)\left(x^2+kx+2*12^2\right)$

$= x^4 + 4*12^2x^2 -k^2x^2 +4*12^4 \Rightarrow k= \pm 24$

$\frac{\left(x^2+24+288\right)\left(x^2-24x+288\right)}{x^2+24x+288} = 144x-288$

$x^2-24x+288=144x-288$

$x^2 -168x + 576$

$x_1 + x_2 = \frac{168*2}{2} = 168$

By the Sophie Germain Identity, we find that $\begin{aligned} x^4 + 4 \cdot 12 &= (x^2 + 2 \cdot 12^2 + 2 \cdot 12x)(x^2 + 2 \cdot 12^2 - 2 \cdot 12x) \\ &= (x^2 + 24x + 288)(x^2 - 24x + 288). \end{aligned}$ Plugging this into our equation, we see that $\frac{x^4 + 4 \cdot 12^4}{x^2 + 24x + 288} = 144x - 288 \\ x^2 - 24x + 288 = 144x - 288 \\ x^2 - 168x + 2 \cdot 288 = 0.$ Therefore, our equation has two roots, and by Vieta's Formulae, we can see that their sum is equal to $\boxed{168}$ .

Firstly have a look at this

Using Sophie Germain Identity we can factorize the numerator in this way:

$x^{4} + 4*12^{4} = (x^{2} + 288 + 24x)(x^{2} + 288 - 24x)$

Now we can simplify it with the first factor of the denominator and so we get:

$x^{2} + 288 - 24x = 144x - 288$

Simplifying:

$x^2 - 168x + 576 = 0$

Using the formula we get as a solutions:

$\frac{168 \pm \sqrt{25920}} {2}$

Adding both solutions we get 168

By Sophie Germain Identity,

$x^4+4\cdot 12^4=(x^2-24x+288)(x^2+24x+288)$

Hence, the above equation can be reduced into

$x^2-24x+288=144x-288$

which is equivalent to

$x^2-168x+576=0$

By Vieta's Formula, the sum of roots = 168.