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Question

$\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(a x-b)^2+(b x-c)^2=0 $

Vedclass · Accepted Answer

$36$

Vedclass · Answer

$\left(a^2+b^2 ight) 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(a x-b)^2+(b x-c)^2=0 $ $ \Rightarrow a x-b=0, \quad b x-c=0 $ $ \Rightarrow a+b>c \quad b+c>a \quad \quad c+a>b$ $a+a x>b x $ $a+a x > a x^2 $ $ x^2-x-1 < 0$ $a x+b x > a $ $ a x+a x^2 > a $ $ x^2+x-1 > 0$ $a x^2+a>a x$ $x^2-x+1>0$ always true $\frac{1-\sqrt{5}}{2}<\mathrm{x}<\frac{1+\sqrt{5}}{2}$ $\mathrm{x}<\frac{-1-\sqrt{5}}{2}, ext { or } \mathrm{x}>\frac{-1+\sqrt{5}}{2} $ $\Rightarrow \frac{\sqrt{5}-1}{2} < x < \frac{\sqrt{5}+1}{2}$ $\Rightarrow \alpha=\frac{\sqrt{5}-1}{2}, \beta=\frac{\sqrt{5}+1}{2}$ $12\left(\alpha^2+\beta^2 ight)=12\left(\frac{(\sqrt{5}-1)^2+(\sqrt{5}+1)^2}{4} ight)=36$

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