Let a,b,c be the length of three sides of a triangle satisfying the condition (a^2+b^2)x^2-2b(a+c) x+(b^2+ ^2)=0. If the set of all possible values…

Given,

$$({a}^{2}+{b}^{2}){x}^{2}-2b(a+c)x+{b}^{2}+{c}^{2}=0$$(a2+b2)x2−2b(a+c)x+b2+c2=0

$$\Rightarrow {a}^{2}{x}^{2}-2abx+{b}^{2}+{b}^{2}{x}^{2}-2bcx+{c}^{2}=0$$⇒a2x2−2abx+b2+b2x2−2bcx+c2=0

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

$$\Rightarrow ax-b=0,bx-c=0$$⇒ax−b=0,bx−c=0

$$\Rightarrow ax=b,bx=c$$⇒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 $a+b>c$a+b>c we get,

$$\Rightarrow a+ax>bx$$⇒a+ax>bx

$$\Rightarrow a+ax>a{x}^{2}$$⇒a+ax>ax2

$$\Rightarrow {x}^{2}-x-1<0$$⇒x2−x−1<0

$$\Rightarrow \frac{1-\sqrt{5}}{2}<x<\frac{1+\sqrt{5}}{2}...(i)$$⇒21−5​​<x<21+5​​...(i)

Similarly, for $b+c>a$b+c>a we get,

$$\Rightarrow {x}^{2}+x-1>0$$⇒x2+x−1>0

$$\Rightarrow x\in (-\infty ,\frac{-1-\sqrt{5}}{2})\cup (\frac{-1+\sqrt{5}}{2},\infty )....(ii)$$⇒x∈(−∞,2−1−5​​)∪(2−1+5​​,∞)....(ii)

And for $c+a>b$c+a>b we get,

$$\Rightarrow {x}^{2}-x+1>0$$⇒x2−x+1>0

$$\Rightarrow x\in R........(iii)$$⇒x∈R........(iii)

So, from the equation (i),(ii)&(iii) we get,

$$x\in (\frac{-1+\sqrt{5}}{2},\frac{1+\sqrt{5}}{2})$$x∈(2−1+5​​,21+5​​)

$$\Rightarrow \alpha =\frac{\sqrt{5}-1}{2},\beta =\frac{\sqrt{5}+1}{2}$$⇒α=25​−1​,β=25​+1​

Hence, $12({\alpha }^{2}+{\beta }^{2})=12(\frac{{(\sqrt{5}-1)}^{2}+{(\sqrt{5}+1)}^{2}}{4})=36$12(α2+β2)=12(4(5​−1)2+(5​+1)2​)=36