Question

Let \(a, b, c\) be the length of three sides of a triangle satisfying the condition \(\left(a^{2}+b^{2}\right) x^{2}-2 b(a+c)x+\left(b^{2}+c^{2}\right)=0\). If the set of all possible values of \(\mathrm{x}\) is the interval \((\alpha, \beta)\), then \(12\left(\alpha^{2}+\beta^{2}\right)\) is equal to _____.

Answer 1

Correct answer: 36

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

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

\(x = \frac ba, x = \frac cb\)

Now

\(\mathrm{b}^2=\mathrm{ac}\)

\(|\mathrm{a}-\mathrm{c}|<\mathrm{b}<\mathrm{a}+\mathrm{c} \quad\quad\mathrm{b}^2=\mathrm{ac} \times \frac{\mathrm{a}}{\mathrm{a}}\)

\(\left|1-\frac{\mathrm{c}}{\mathrm{a}}\right|<\frac{\mathrm{b}}{\mathrm{a}}<1+\frac{\mathrm{c}}{\mathrm{a}} \quad \quad\frac{\mathrm{b}^2}{\mathrm{a}^2}=\frac{\mathrm{c}}{\mathrm{a}} \)

\(\left|1-\frac{\mathrm{c}}{\mathrm{a}}\right|<\mathrm{x}<1+\frac{\mathrm{c}}{\mathrm{a}} \quad\quad\mathrm{x}^2=\frac{\mathrm{c}}{\mathrm{a}} \)

\(\left|1-\mathrm{x}^2\right|<\mathrm{x}<1+\mathrm{x}^2 \)

Case I: \(\mathrm{x}<1+\mathrm{x}^2\)

always +ve

Case II: \(\left|1-x^2\right|<x\)

\(-\mathrm{x}<1-\mathrm{x}^2<\mathrm{x}\)

Now

\(1-x^2<x \)

\(x^2+x-1>0\)

\( x=\frac{-1 \pm \sqrt{5}}{2}\)

Now

\(-x<1-x^2\)

\(x^2-x-1<0 \)

\(x=\frac{1 \pm \sqrt{5}}{2}\)

\( \Rightarrow \alpha=\frac{-1 +\sqrt{5}}{2}, \beta=\frac{1+\sqrt{5}}{2}\)

Now

\(12\left(\alpha^{2}+\beta^{2}\right) = 12 \left( \left(\frac{-1 + \sqrt 5}2 \right)^2 + \left(\frac{1 + \sqrt 5}2 \right)^2 \right)\)

\(=12\left(\frac{(-1+\sqrt{5})^{2}+(1+\sqrt{5})^{2}}{4}\right)\)

\(=36\)