### Core Logic
Given equation:
(a^2+b^2)x^2 - 2b(a+c)x + b^2+c^2 = 0$(a^2+b^2)x^2 - 2b(a+c)x + b^2+c^2 = 0$.
Expand and rearrange into perfect squares:
(a^2x^2 - 2abx + b^2) + (b^2x^2 - 2bcx + c^2) = 0$$(a^2x^2 - 2abx + b^2) + (b^2x^2 - 2bcx + c^2) = 0$$
(ax - b)^2 + (bx - c)^2 = 0$$(ax - b)^2 + (bx - c)^2 = 0$$
Since squares must be non-negative, each term is zero:
ax - b = 0 implies x = fracba$$ax - b = 0 \implies x = \frac{b}{a}$$
bx - c = 0 implies x = fraccb$$bx - c = 0 \implies x = \frac{c}{b}$$
Thus,
b = ax$b = ax$ and
c = bx = ax^2$c = bx = ax^2$.
Since
a,b,c$a,b,c$ form a triangle, the
triangle inequality holds:
1)
a + b > c implies a + ax > ax^2 implies x^2 - x - 1 < 0 implies frac1-sqrt52 < x < frac1+sqrt52$a + b > c \implies a + ax > ax^2 \implies x^2 - x - 1 < 0 \implies \frac{1-\sqrt{5}}{2} < x < \frac{1+\sqrt{5}}{2}$
2)
a + c > b implies a + ax^2 > ax implies x^2 - x + 1 > 0$a + c > b \implies a + ax^2 > ax \implies x^2 - x + 1 > 0$ (Always true for real
x$x$)
3)
b + c > a implies ax + ax^2 > a implies x^2 + x - 1 > 0 implies x > frac-1+sqrt52$b + c > a \implies ax + ax^2 > a \implies x^2 + x - 1 > 0 \implies x > \frac{-1+\sqrt{5}}{2}$ or
x < frac-1-sqrt52$x < \frac{-1-\sqrt{5}}{2}$.
Taking the intersection (and noting
x > 0$x > 0$ since sides are positive):
fracsqrt5-12 < x < fracsqrt5+12$$\frac{\sqrt{5}-1}{2} < x < \frac{\sqrt{5}+1}{2}$$
So,
alpha = fracsqrt5-12$\alpha = \frac{\sqrt{5}-1}{2}$ and
beta = fracsqrt5+12$\beta = \frac{\sqrt{5}+1}{2}$.
Calculate
12(alpha^2 + beta^2)$12(\alpha^2 + \beta^2)$:
12 left( frac6-2sqrt54 + frac6+2sqrt54 right) = 12 left( frac124 right) = 36$$12 \left( \frac{6-2\sqrt{5}}{4} + \frac{6+2\sqrt{5}}{4} \right) = 12 \left( \frac{12}{4} \right) = 36$$
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Maths: Complex Numbers and Quadratic Equations
Class 11 Maths: Straight Lines
