JEE Main 2024 — 31 January, Shift 2 — Question 20

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

$\Rightarrow a^{2} x^{2}-2 a b x+b^{2}+b^{2} x^{2}-2 b c x+c^{2}=0$

$\Rightarrow(\mathrm{ax}-\mathrm{b})^{2}+(\mathrm{bx}-\mathrm{c})^{2}=0$

$\Rightarrow \mathrm{ax}-\mathrm{b}=0, \mathrm{bx}-\mathrm{c}=0$

$\Rightarrow \mathrm{a}+\mathrm{b}>\mathrm{c} ,\quad \mathrm{b}+\mathrm{c}>\mathrm{a}, \quad \mathrm{c}+\mathrm{a}>\mathrm{b}$

$\mathrm{a}+\mathrm{ax}>\mathrm{bx},\quad \mathrm{ax}+\mathrm{bx}>\mathrm{a},\quad \mathrm{ax}^{2}+\mathrm{a}>\mathrm{ax}$

$a+a x>x^{2} ,\quad a x+a x^{2}>a ,\quad x^{2}-x+1>0$

$\mathrm{x}^{2}-\mathrm{x}-1<0 ,\qquad \mathrm{x}^{2}+\mathrm{x}-1>0$

always true $\frac{1-\sqrt{5}}{2}<\mathrm{x}<\frac{1+\sqrt{5}}{2}$

$\mathrm{x}<\frac{-1-\sqrt{5}}{2}$, or $\mathrm{x}>\frac{-1+\sqrt{5}}{2}$

$$\begin{array}{l} \Rightarrow \frac{{\sqrt 5 - 1}}{2} < x < \frac{{\sqrt 5 + 1}}{2}\\ \Rightarrow \alpha = \frac{{\sqrt 5 - 1}}{2},\beta = \frac{{\sqrt 5 + 1}}{2}\\12\left( {{\alpha ^2} + {\beta ^2}} \right) = 12\left( {\frac{{{{\left( {\sqrt 5 - 1} \right)}^2} + {{\left( {\sqrt 5 + 1} \right)}^2}}}{4}} \right) = 36\end{array}$$