Question

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

Answer 1

Correct Answer: 36

Given: 

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

\[ (ax - b)^2 + (bx - c)^2 = 0 \]

\[ x = \frac{b \pm \sqrt{b^2 - ac}}{a^2 + b^2} \]

Step 1:

Consider the triangle with sides \( a, b, c \): \[ |h - k| < c < (a + b) \]

Since \( b^2 = ac \), \[ |a - c| < b < a + c \] Substituting \( b^2 = ac \): \[ |1 - \xi| < \xi < 1 + \xi \]

Hence, \[ |1 - \xi| < \xi < 1 + \xi \] \[ |1 - x^2| < x < 1 + x^2 \]

Case 1: \( b < x < 1 + x^2 \)

This is always positive.

Case 2: \( |1 - x^2| < x \)

\[ -x < 1 - x^2 < x \]

Now:

\[ 1 - x^2 < x \] \[ x^2 + x - 1 > 0 \]

Solving: \[ x = \frac{-1 \pm \sqrt{5}}{2} \]

\[ -x < 1 - x^2 \] \[ x^2 - x - 1 < 0 \]

Hence, \[ x = \frac{1 \pm \sqrt{5}}{2} \]

Now:

\[ \alpha = \frac{-1 + \sqrt{5}}{2}, \quad \beta = \frac{-1 - \sqrt{5}}{2} \]

Step 2:

\[ 12(a^2 + b^2) = 12 \left[\left(\frac{-1 + \sqrt{5}}{2}\right)^2 + \left(\frac{-1 - \sqrt{5}}{2}\right)^2\right] \]

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

\[ = 12 \times \frac{12}{4} = 36 \]

∴ Final Answer:

\[ \boxed{36} \]

Answer 2

The given quadratic equation in \( x \) is:

\[ (a^2 + b^2)x^2 - 2b(a + c)x + (b^2 + c^2) = 0. \]

This can be written in the form:

\[ (ax - b)^2 + (bx - c)^2 = 0. \]

Thus, we deduce that the discriminant must satisfy conditions related to triangle inequalities, leading us to intervals of \( x \) values.

By evaluating the possible values of \( x \), we find that the interval \((\alpha, \beta)\) corresponds to:

\[ \alpha = \frac{1 - \sqrt{5}}{2}, \quad \beta = \frac{1 + \sqrt{5}}{2}. \]

Then, calculate \(12(\alpha^2 + \beta^2)\):

\[ 12(\alpha^2 + \beta^2) = 36. \]

Thus, the answer is:

\[ 36. \]