Question Type: Quadratic Equations

Step-by-step solution:

If two quadratic equations have both roots common, then the ratio of the corresponding coefficients must be equal.

Let the quadratic equations be:

\(2x^2 + Kx + 8 = 0\) ...(1)

\(3x^2 + 4x + 12 = 0\) ...(2)

For both equations to have common roots, the ratio of their corresponding coefficients must be equal. Therefore:

\(\frac{2}{3} = \frac{K}{4} = \frac{8}{12}\)

We can use any two ratios to solve for K. Let's use the first and third ratios:

\(\frac{2}{3} = \frac{8}{12}\)

This simplifies to \(\frac{2}{3} = \frac{2}{3}\), which is true, confirming that the equations could potentially have common roots.

Now let's use the first and second ratios to solve for K:

\(\frac{2}{3} = \frac{K}{4}\)

Cross-multiplying, we get:

\(2 \times 4 = 3 \times K\)

\(8 = 3K\)

\(K = \frac{8}{3}\)

Therefore, the value of K is \(\frac{8}{3}\).

Eliminating Incorrect Options:

• Options 2, 3, and 4 are incorrect because they do not satisfy the condition of equal ratios of corresponding coefficients for the two quadratic equations to have common roots.