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

Quadratic Equations and Inequalities — JEE Main · 2024 · 31 Jan 2024 · Shift 2 Question with Solution

Let

a, b, c

be the length 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 ______.

Answer: 36

The condition

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

can be rewritten as a sum of two squares:

(a x - b)^{2} + (b x - c)^{2} = a^{2} x^{2} - 2 ab x + b^{2} + b^{2} x^{2} - 2 b c x + c^{2} = (a^{2} + b^{2}) x^{2} - 2 b (a + c) x + (b^{2} + c^{2})

,

which matches exactly. Since a sum of two real squares is zero only if each square is zero,

a x = b

and

b x = c

, i.e.

b = a x

c = a x^{2}

.

Since

a, b, c

are sides of a triangle, they must satisfy the triangle inequalities; substituting

b = a x, c = a x^{2}

and dividing by

a > 0

a + b > c \Rightarrow 1 + x > x^{2} \Rightarrow x^{2} - x - 1 < 0 \Rightarrow \frac{1 - 5}{2} < x < \frac{1 + 5}{2}

b + c > a \Rightarrow x + x^{2} > 1 \Rightarrow x^{2} + x - 1 > 0 \Rightarrow x < \frac{- 1 - 5}{2}

x > \frac{- 1 + 5}{2}

c + a > b \Rightarrow x^{2} + 1 > x \Rightarrow x^{2} - x + 1 > 0

, which always holds (discriminant

< 0

).

Intersecting the first two conditions leaves

x \in (\frac{5 - 1}{2}, \frac{5 + 1}{2})

, so

α = \frac{5 - 1}{2}, β = \frac{5 + 1}{2}

α^{2} = \frac{3 - 5}{2}, β^{2} = \frac{3 + 5}{2}

, so

α^{2} + β^{2} = 3

12 (α^{2} + β^{2}) = 36

JEE Main · 2024 · 31 Jan 2024 · Shift 2 · Integer answer