Problem 14

We have $\displaystyle (x^2+2)^2+8x=6x(x^2+2)$ .

Rearranging gives us $\displaystyle (x^2+2)^2-6x(x^2+2)+8x^2=0$

Solving it as a Quadratic in $\displaystyle (x^2+2)$ , by the Sridharacharya's Rules gives- $\displaystyle (x^2+2)=\dfrac{6x \pm \sqrt{36x^2-32x^2}}{2}$

Simplifying, we get $\displaystyle x^2-4x+2=0$ or $\displaystyle x^2-2x+2=0$ .

The former gives us $\displaystyle 2$ real roots adding up, by Vieta's Relation, to $\displaystyle 4$ , and the latter has no real roots.

Thus, the sum of the real roots of the given equation is $\displaystyle \boxed{4}$ .

$\left( x^2+2 \right)^2+8x^2 = 6x\left( x^2+2 \right)$

$\begin{aligned} \Rightarrow \left( x^2+2 \right)^2 - 6x \left( x^2+2 \right) + 8x^2 & = 0 \\ \left( (x^2+2) - 2x \right) \left( (x^2+2) - 4x \right) & = 0 \\ \left( x^2-2x+2 \right) \left( x^2-4x+2 \right) & = 0 \end{aligned}$

$\Rightarrow \begin{cases} x^2-2x+2 = 0 & \Rightarrow x = 1\pm i \\ x^2-4x+2 = 0 & \Rightarrow x = 2 \pm \sqrt{2} \end{cases}$

Therefore, the sum of real roots are $2 + \sqrt{2} +2 - \sqrt{2} = \boxed{4}$

Considere $x^2 + 2 = \alpha$ , segue que,

$\alpha^2 + 8x^2 = 6x \cdot \alpha\\\\ \alpha^2 - 6x \cdot \alpha + 8x^2 = 0 \\\\ (\alpha - 2x)(\alpha - 4x) = 0 \\\\ \underbrace{(x^2 - 2x + 2)}_{\Delta < 0} \cdot \underbrace{(x^2 - 4x + 2)}_{\Delta > 0} = 0$

Daí, deve-se considerar apenas as raízes da equação $x^2 - 4x + 2$ .

Isto posto,

$\text{Soma} = \frac{- b}{a} \\\\ \text{Soma} = \frac{- (- 4)}{1} \\\\ \boxed{\text{Soma} = 4}$

Let $m=x^2+2$

$\Rightarrow (x^2+2)^2+8x^2-6x(x^2+2)=m^2-6xm+8x^2$

$m=\frac{6x\pm\sqrt{(-6x)^2-4(8x^2)}}{2}=\frac{6x\pm 2x}{2}=4x,2x$

As $m=x^2+2$

$x^2+2=4x\Rightarrow x^2+2-4x=0$ this solutions are $x=2-\sqrt{2},2+\sqrt{2}$

$x^2+2=2x\Rightarrow x^2+2-2x=0$ this one has complex solutions

Therefore the sum of real root are $2-\sqrt{2}+2+\sqrt{2}=\boxed{4}$