Let a, b, c be the length of three sides of a triangle satisfying the condition (a² + b²)x² - 2b(a +

Q: Wait, why can we assume $(ax - b)^2 + (bx - c)^2 = 0$ implies both terms are zero?

That is a fantastic question! Remember that $a, b, c,$ and $x$ are all real numbers. The square of any real number is always non-negative (greater than or equal to zero). If you have two non-negative numbers that add up to zero, the only mathematical possibility is that each of them must be zero individually. It is a powerful property of real numbers that simplifies complex equations instantly!

Q: Why do we need to check all three triangle inequalities?

The triangle inequality is the fundamental definition of a triangle's existence. It states that the sum of any two sides must be strictly greater than the third side. We must satisfy $a + b > c$, $b + c > a$, and $a + c > b$ simultaneously. If we only checked one, we might include values of $x$ that don't actually form a valid triangle. We must be rigorous to ensure our solution is physically possible.

Q: How do we know $x$ is positive?

Great observation! Since $x$ is the ratio of two side lengths ($x = b/a$ and $x = c/b$), and side lengths are always positive, their ratio must also be positive. This is why we can safely ignore any negative roots or intervals when solving our inequalities.

Question

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

Sigma Percentile · Accepted Answer

### Analyzing the Setup Imagine you are standing before a complex, intimidating quadratic equation: $$(a^2 + b^2)x^2 - 2b(a + c)x + (b^2 + c^2) = 0$$ At first glance, it looks like a mess of variables. But in the world of JEE Advanced, equations are rarely just random collections of symbols; they are puzzles waiting for the right perspective. Let us embark on a journey to decode this. ### The Algebraic Revelation The first step is to peel back the layers. Let us expand the equation: $$a^2x^2 + b^2x^2 - 2abx - 2bcx + b^2 + c^2 = 0$$ Now, look at the terms and rearrange them to reveal hidden structures: $$(a^2x^2 - 2abx + b^2) + (b^2x^2 - 2bcx + c^2) = 0$$ Suddenly, the chaos transforms into a beautiful, recognizable pattern. These are perfect squares! The equation collapses into: $$(ax - b)^2 + (bx - c)^2 = 0$$ This is the moment of clarity. Since the sum of two squares of real numbers is zero, each square must be zero individually. Thus, $ax - b = 0$ and $bx - c = 0$. This leads us to $x = \frac{b}{a}$ and $x = \frac{c}{b}$. Equating these, we find $\frac{b}{a} = \frac{c}{b}$, which implies $b^2 = ac$. Our triangle sides are in a **Geometric Progression**. ### The Geometric Heartbeat Now, we must respect the physical reality of our triangle. The **triangle inequality** is our guiding star. We know that for any triangle with sides $a, b, c$, the sum of any two sides must be greater than the third. Since $b = ax$ and $c = ax^2$, we can express all sides in terms of $a$ and $x$. The first inequality, $a + b > c$, becomes $a + ax > ax^2$. Dividing by $a$ (which is positive), we get $1 + x > x^2$, or $x^2 - x - 1 < 0$. Solving this quadratic inequality gives us: $$x < \frac{1 + \sqrt{5}}{2}$$ The second inequality, $b + c > a$, becomes $ax + ax^2 > a$, which simplifies to $x^2 + x - 1 > 0$. This gives us: $$x > \frac{\sqrt{5} - 1}{2}$$ The third inequality, $a + c > b$, leads to $x^2 - x + 1 > 0$, which is always true for all real $x$ because its discriminant is negative. ### The Final Synthesis We have our bounds! The set of all possible values for $x$ is the interval $(\alpha, \beta)$, where: $$\alpha = \frac{\sqrt{5} - 1}{2}, \quad \beta = \frac{\sqrt{5} + 1}{2}$$ The problem asks us to calculate $12(\alpha^2 + \beta^2)$. Let us compute the squares: $$\alpha^2 = \frac{6 - 2\sqrt{5}}{4}, \quad \beta^2 = \frac{6 + 2\sqrt{5}}{4}$$ Adding them together, the irrational terms cancel out perfectly: $$\alpha^2 + \beta^2 = \frac{6 - 2\sqrt{5} + 6 + 2\sqrt{5}}{4} = \frac{12}{4} = 3$$ Finally, $12 imes 3 = 36$. It is a moment of pure mathematical elegance. The final result is **36**.

Enter Numerical Value:

Visualized Solution

The Given Equation

Expanding the Terms

Forming Perfect Squares

Solving for $x$

Triangle Inequality 1

Solving the First Inequality

Triangle Inequality 2

Solving the Second Inequality

Triangle Inequality 3

Intersection of Conditions

Calculating $α^{2} + β^{2}$

Final Answer

The Sigma Insight: Solution of Quadratic Equations

Analyzing the Setup

The Algebraic Revelation

The Geometric Heartbeat

The Final Synthesis

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