Let a , b , c be the lengths of three sides of a triangle satisfying the condition ( a 2 + b 2 ) x 2 - 2 b ( a + c ) x + ( b 2 + c 2 ) = 0 . If the set of all possible values of x is the interval, then is equal to _____ . [2024]

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Solution

Q.
Let $a, b, c$ be the lengths of three sides of a triangle satisfying the condition $(a^{2} + b^{2}) x^{2} - 2 b (a + c) x + (b^{2} + c^{2}) = 0 .$ If the set of all possible values of $x$ is the interval, $(α, β)$ then $12 (α^{2} + β^{2})$ is equal to _____ . [2024]

Ans.
(36)

We have, $(a^{2} + b^{2}) x^{2} - 2 b (a + c) x + (b^{2} + c^{2}) = 0$

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

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

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

$\Rightarrow a x - b = 0, b x - c = 0$

$\Rightarrow a x = b, b x = c \Rightarrow a x^{2} = b x, b x^{2} = c x$

Since, $a + b > c, b + c > a and c + a > b$

$a + a x > b x, a x + b x > a and b x + a > a x$

$\Rightarrow a + a x > a x^{2}, a x + a x^{2} > a and a x^{2} + a > a x$

$\Rightarrow x^{2} - x - 1 < 0, x^{2} + x - 1 > 0 and x^{2} - x + 1 > 0$

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

$\Rightarrow \frac{\sqrt{5} - 1}{2} < x < \frac{\sqrt{5} + 1}{2} \Rightarrow α = \frac{\sqrt{5} - 1}{2}, β = \frac{\sqrt{5} + 1}{2}$

$∴$ $12 (α^{2} + β^{2}) = 36$

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