Question:

**Step 1: Identify the roots and apply Vieta's formulas.** Let the roots of $ax^2 + bx + c = 0$ be $\alpha$ and $\beta$. By Vieta's formulas, the sum of the roots is $\alpha + \beta = -\frac{b}{a}$, and the product of the roots is $\alpha\beta = \frac{c}{a}$. **Step 2: Calculate the Arithmetic Mean $x_1$.** The arithmetic mean of the roots is: $$x_1 = \frac{\alpha + \beta}{2} = \frac{-b/a}{2} = -\frac{b}{2a}$$ **Step 3: Calculate the Harmonic Mean $x_2$.** The harmonic mean of two numbers $p$ and $q$ is $\frac{2pq}{p+q}$. Thus, for the roots $\alpha$ and $\beta$: $$x_2 = \frac{2\alpha\beta}{\alpha + \beta} = \frac{2(c/a)}{(-b/a)} = \frac{2c}{-b} = -\frac{2c}{b}$$ **Step 4: Determine the sum of $x_1$ and $x_2$.** The sum of the new roots is: $$x_1 + x_2 = -\frac{b}{2a} + \left(-\frac{2c}{b}\right) = -\frac{b}{2a} - \frac{2c}{b}$$ To combine these terms, find a common denominator $2ab$: $$x_1 + x_2 = \frac{-b^2 - 4ac}{2ab} = -\frac{b^2 + 4ac}{2ab}$$ **Step 5: Determine the product of $x_1$ and $x_2$.** The product of the new roots is: $$x_1 \cdot x_2 = \left(-\frac{b}{2a}\right) \cdot \left(-\frac{2c}{b}\right) = \frac{b \cdot 2c}{2ab} = \frac{c}{a}$$ **Step 6: Construct the quadratic equation.** The quadratic equation whose roots are $x_1$ and $x_2$ is given by $x^2 - (x_1 + x_2)x + x_1x_2 = 0$. Substituting the sum and product found in Steps 4 and 5: $$x^2 - \left(-\frac{b^2 + 4ac}{2ab}\right)x + \frac{c}{a} = 0$$ $$x^2 + \frac{b^2 + 4ac}{2ab}x + \frac{c}{a} = 0$$ To eliminate the denominators, multiply the entire equation by $2ab$: $$2abx^2 + (b^2 + 4ac)x + 2bc = 0$$