Let a, b, c be the length of three sides of a triangle satisfying...

Given,

$\left({a}^{2} + {b}^{2}\right){x}^{2} - 2b\left(a + c\right)x + {b}^{2} + {c}^{2} = 0$

$⇒ {a}^{2}{x}^{2} - 2abx + {b}^{2} + {b}^{2}{x}^{2} - 2bcx + {c}^{2} = 0$

$⇒ {\left(ax - b\right)}^{2} + {\left(bx - c\right)}^{2} = 0$

$⇒ ax - b = 0, bx - c = 0$

$⇒ ax = b, bx = c$

Now, we know that the sum of two sides is always greater than the third side of a triangle,

Now, taking $a + b > c$ we get,

$⇒ a + ax > bx$

$⇒ a + ax > a{x}^{2}$

$⇒ {x}^{2} - x - 1 < 0$

$⇒ \dfrac{1 - \sqrt{5}}{2} < x < \dfrac{1 + \sqrt{5}}{2} . . . \left(i\right)$

Similarly, for $b + c > a$ we get,

$⇒ {x}^{2} + x - 1 > 0$

$⇒ x \in \left(- \infty , \dfrac{- 1 - \sqrt{5}}{2}\right) \cup \left(\dfrac{- 1 + \sqrt{5}}{2}, \infty\right) . . . . \left(ii\right)$

And for $c + a > b$ we get,

$⇒ {x}^{2} - x + 1 > 0$

$⇒ x \in R . . . . . . . . \left(iii\right)$

So, from the equation \left(i\right), \left(ii\right) \text{ and } \left(iii\right)$we get,</p><p>$x \in \left(\dfrac{- 1 + \sqrt{5}}{2}, \dfrac{1 + \sqrt{5}}{2}\right)$</p><p>$⇒ \alpha = \dfrac{\sqrt{5} - 1}{2}, \beta = \dfrac{\sqrt{5} + 1}{2}$</p><p>Hence,$12\left({\alpha}^{2} + {\beta}^{2}\right) = 12\left(\dfrac{{\left(\sqrt{5} - 1\right)}^{2} + {\left(\sqrt{5} + 1\right)}^{2}}{4}\right) = 36$