Key Concepts and Formulas

• Vieta's Formulas: For a quadratic equation $ax^2 + bx + c = 0$, the sum of the roots is $-b/a$ and the product of the roots is $c/a$.
• Quadratic Formula: For a quadratic equation $ax^2 + bx + c = 0$, the roots are given by $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
• Divisors: Understanding the divisors of a number is crucial for identifying potential integer roots.

Step-by-Step Solution

Step 1: Define the Equations and Roots

Let the first quadratic equation be $x^2 - 6x + a = 0 \quad (1)$ Let the second quadratic equation be $x^2 - cx + 6 = 0 \quad (2)$ Let $\alpha$ be the common root. Let $r_1$ be the other root of equation (1), and $r_2$ be the other root of equation (2).

Step 2: Apply Vieta's Formulas to Equation (1)

The sum of the roots of equation (1) is $\alpha + r_1 = 6$. The product of the roots of equation (1) is $\alpha r_1 = a$.

Step 3: Apply Vieta's Formulas to Equation (2)

The sum of the roots of equation (2) is $\alpha + r_2 = c$. The product of the roots of equation (2) is $\alpha r_2 = 6$.

Step 4: Express $r_1$ and $r_2$ in terms of $\alpha$

From the sum of roots of equation (1), we have $r_1 = 6 - \alpha$. From the product of roots of equation (2), we have $r_2 = \frac{6}{\alpha}$. Since $r_2$ is an integer, $\alpha$ must be a divisor of 6. Thus, $\alpha \in \{\pm 1, \pm 2, \pm 3, \pm 6\}$.

Step 5: Consider the Ratio $r_1 : r_2 = 4 : 3$

If $\frac{r_1}{r_2} = \frac{4}{3}$, then $3r_1 = 4r_2$. Substitute the expressions for $r_1$ and $r_2$: $3(6 - \alpha) = 4\left(\frac{6}{\alpha}\right)$ $18 - 3\alpha = \frac{24}{\alpha}$ Multiply by $\alpha$ (since $\alpha \neq 0$): $18\alpha - 3\alpha^2 = 24$ $3\alpha^2 - 18\alpha + 24 = 0$ Divide by 3: $\alpha^2 - 6\alpha + 8 = 0$ $(\alpha - 2)(\alpha - 4) = 0$ So, $\alpha = 2$ or $\alpha = 4$.

Step 6: Check the Integer Root Condition for $\alpha = 2$ and $\alpha = 4$

• If $\alpha = 2$, then $r_1 = 6 - 2 = 4$ and $r_2 = \frac{6}{2} = 3$. Both $r_1$ and $r_2$ are integers, and $\frac{r_1}{r_2} = \frac{4}{3}$.
• If $\alpha = 4$, then $r_1 = 6 - 4 = 2$ and $r_2 = \frac{6}{4} = \frac{3}{2}$. $r_2$ is not an integer, so $\alpha = 4$ is not a valid solution.

Step 7: Consider the Ratio $r_1 : r_2 = 3 : 4$

If $\frac{r_1}{r_2} = \frac{3}{4}$, then $4r_1 = 3r_2$. Substitute the expressions for $r_1$ and $r_2$: $4(6 - \alpha) = 3\left(\frac{6}{\alpha}\right)$ $24 - 4\alpha = \frac{18}{\alpha}$ Multiply by $\alpha$: $24\alpha - 4\alpha^2 = 18$ $4\alpha^2 - 24\alpha + 18 = 0$ Divide by 2: $2\alpha^2 - 12\alpha + 9 = 0$ Using the quadratic formula: $\alpha = \frac{12 \pm \sqrt{144 - 72}}{4} = \frac{12 \pm \sqrt{72}}{4} = \frac{6 \pm 3\sqrt{2}}{2}$ These values are irrational. Thus, $r_1$ and $r_2$ would also be irrational, violating the integer root condition.

Step 8: Verify the Solution $\alpha = 2$

If $\alpha = 2$, then $r_1 = 4$ and $r_2 = 3$. Equation (1) becomes $x^2 - 6x + a = 0$, where $a = \alpha r_1 = 2 \cdot 4 = 8$. The equation is $x^2 - 6x + 8 = 0$, which has roots 2 and 4. Equation (2) becomes $x^2 - cx + 6 = 0$, where $c = \alpha + r_2 = 2 + 3 = 5$. The equation is $x^2 - 5x + 6 = 0$, which has roots 2 and 3. The common root is 2, and the other roots are 4 and 3, which are integers in the ratio 4:3.

Common Mistakes & Tips

• Forgetting to check integer root condition: After finding potential values for the common root, ensure that the other roots are integers.
• Incorrectly assigning the ratio: Consider both possible assignments of the ratio (4:3 and 3:4) to ensure all possibilities are explored.
• Assuming $\alpha \ne 0$ without justification: While true in this case as $\alpha r_2 = 6$, explicitly state why $\alpha$ cannot be zero.

Summary

By applying Vieta's formulas and considering the given conditions, we found that the common root of the two quadratic equations is 2. This value satisfies the conditions of integer roots and the specified ratio.

The final answer is \boxed{2}, which corresponds to option (D).