Key Concepts and Formulas

• Quadratic Equation and Discriminant: For a quadratic equation $ax^2 + bx + c = 0$, the discriminant is $\Delta = b^2 - 4ac$. The nature of roots depends on $\Delta$.
• Sum of Squares Property: If $A^2 + B^2 = 0$, where $A$ and $B$ are real numbers, then $A = 0$ and $B = 0$.
• Triangle Inequality: For a triangle with sides $a, b, c$, the following inequalities must hold: $a + b > c$, $b + c > a$, and $c + a > b$.

Step-by-Step Solution

Step 1: Rewriting the Quadratic Equation as a Sum of Squares

We are given the quadratic equation $(a^2+b^2) x^2-2 b(a+c) x+\left(b^2+c^2\right)=0$. Our goal is to rewrite this equation in a form that allows us to use the sum of squares property. Expanding the middle term, we have: $a^2x^2 + b^2x^2 - 2abx - 2bcx + b^2 + c^2 = 0$ Rearranging the terms, we can group them to form perfect squares: $(a^2x^2 - 2abx + b^2) + (b^2x^2 - 2bcx + c^2) = 0$ This simplifies to: $(ax - b)^2 + (bx - c)^2 = 0$ Since the sum of squares is equal to zero, each term must be zero.

Step 2: Deriving Relationships Between Side Lengths and x

From the previous step, we have: $ax - b = 0 \quad \text{and} \quad bx - c = 0$ Solving for $x$ in each equation, we get: $x = \frac{b}{a} \quad \text{and} \quad x = \frac{c}{b}$ Since both expressions must be equal, we have: $\frac{b}{a} = \frac{c}{b}$ Cross-multiplying, we get the relationship: $b^2 = ac$ This means that $a, b, c$ are in Geometric Progression (G.P.). Also, since $a$ and $b$ are lengths, $x = \frac{b}{a} > 0$.

Step 3: Applying the Triangle Inequality

Since $a, b, c$ are the sides of a triangle, they must satisfy the triangle inequality. We will use the relations $b = ax$ and $c = bx$ (or $c = ax^2$) to express the triangle inequality in terms of $a$ and $x$.

1. $a + b > c$ Substituting $b = ax$ and $c = ax^2$: $a + ax > ax^2$ Dividing by $a$ (since $a > 0$): $1 + x > x^2$ $x^2 - x - 1 < 0$

2. $b + c > a$ Substituting $b = ax$ and $c = ax^2$: $ax + ax^2 > a$ Dividing by $a$ (since $a > 0$): $x + x^2 > 1$ $x^2 + x - 1 > 0$

3. $c + a > b$ Substituting $b = ax$ and $c = ax^2$: $ax^2 + a > ax$ Dividing by $a$ (since $a > 0$): $x^2 + 1 > x$ $x^2 - x + 1 > 0$

Step 4: Solving the Quadratic Inequalities

Now we solve the quadratic inequalities for $x$.

1. $x^2 - x - 1 < 0$ The roots of $x^2 - x - 1 = 0$ are $x = \frac{1 \pm \sqrt{5}}{2}$. Since the parabola opens upwards, the inequality is satisfied between the roots: $\frac{1 - \sqrt{5}}{2} < x < \frac{1 + \sqrt{5}}{2}$

2. $x^2 + x - 1 > 0$ The roots of $x^2 + x - 1 = 0$ are $x = \frac{-1 \pm \sqrt{5}}{2}$. Since the parabola opens upwards, the inequality is satisfied outside the roots: $x < \frac{-1 - \sqrt{5}}{2} \quad \text{or} \quad x > \frac{-1 + \sqrt{5}}{2}$

3. $x^2 - x + 1 > 0$ The discriminant of $x^2 - x + 1 = 0$ is $\Delta = (-1)^2 - 4(1)(1) = -3 < 0$. Since the leading coefficient is positive, the inequality is always true for all real $x$.

We need to find the intersection of these solutions and the condition $x > 0$.

The first inequality gives $\frac{1-\sqrt{5}}{2} < x < \frac{1+\sqrt{5}}{2}$. Since $x > 0$, we have $0 < x < \frac{1+\sqrt{5}}{2}$. The second inequality gives $x < \frac{-1-\sqrt{5}}{2}$ or $x > \frac{-1+\sqrt{5}}{2}$. Since $x > 0$, we have $x > \frac{-1+\sqrt{5}}{2}$. Combining these, we have $\frac{-1+\sqrt{5}}{2} < x < \frac{1+\sqrt{5}}{2}$.

Thus, $\alpha = \frac{\sqrt{5}-1}{2}$ and $\beta = \frac{\sqrt{5}+1}{2}$.

Step 5: Calculating $12(\alpha^2 + \beta^2)$

We have $\alpha = \frac{\sqrt{5}-1}{2}$ and $\beta = \frac{\sqrt{5}+1}{2}$. Then, $\alpha^2 = \frac{6 - 2\sqrt{5}}{4} = \frac{3 - \sqrt{5}}{2}$ and $\beta^2 = \frac{6 + 2\sqrt{5}}{4} = \frac{3 + \sqrt{5}}{2}$. Therefore, $\alpha^2 + \beta^2 = \frac{3 - \sqrt{5}}{2} + \frac{3 + \sqrt{5}}{2} = \frac{6}{2} = 3$. Finally, $12(\alpha^2 + \beta^2) = 12(3) = 36$.

Common Mistakes & Tips

• Sign Errors: Be careful with signs when expanding and rearranging terms.
• Triangle Inequality Conditions: Remember to check all three triangle inequalities.
• Domain of x: Don't forget that $x > 0$ because it represents the ratio of side lengths.

Summary

By rewriting the given quadratic equation as a sum of squares, we derived the relationship $b^2 = ac$. Applying the triangle inequality and solving the resulting quadratic inequalities, we found the interval for $x$ to be $(\frac{\sqrt{5}-1}{2}, \frac{\sqrt{5}+1}{2})$. Finally, we calculated $12(\alpha^2 + \beta^2)$ to be 36.

Final Answer

The final answer is \boxed{36}.