Question

The quadratic equations $x^2 - 6x + a = 0$ and $x^2 - cx + 6 = 0$ have one root in common. The other roots of the first and second equation are integers in the ratio 4 : 3. Then common root is ____.

Answer 1

2

Answer 2

Let $\alpha$ be the common root. The roots of $x^2 - 6x + a = 0$ are $\alpha$ and $\beta$. The roots of $x^2 - cx + 6 = 0$ are $\alpha$ and $\gamma$.

From Vieta's formulas:

1. $\alpha + \beta = 6$
2. $\alpha \gamma = 6$

We are given that $\beta$ and $\gamma$ are integers, and their ratio is $4:3$, so $\frac{\beta}{\gamma} = \frac{4}{3}$, which implies $3\beta = 4\gamma$.

From (1), $\beta = 6 - \alpha$. From (2), $\gamma = \frac{6}{\alpha}$ (since $\alpha \neq 0$ as $\alpha\gamma=6$).

Substitute these into the ratio equation: $3(6 - \alpha) = 4\left(\frac{6}{\alpha}\right)$ $18 - 3\alpha = \frac{24}{\alpha}$ Multiplying by $\alpha$: $18\alpha - 3\alpha^2 = 24$ Rearranging into a quadratic equation for $\alpha$: $3\alpha^2 - 18\alpha + 24 = 0$ Dividing by 3: $\alpha^2 - 6\alpha + 8 = 0$ Factoring the quadratic equation: $(\alpha - 2)(\alpha - 4) = 0$ This gives two possible values for the common root: $\alpha = 2$ or $\alpha = 4$.

Case 1: If $\alpha = 2$ $\beta = 6 - 2 = 4$ (integer) $\gamma = 6 / 2 = 3$ (integer) Ratio $\beta : \gamma = 4 : 3$. This case is valid.

Case 2: If $\alpha = 4$ $\beta = 6 - 4 = 2$ (integer) $\gamma = 6 / 4 = 3/2$ (not an integer) This case is invalid.

Therefore, the common root is 2.