Find the value of K if the quadratic equations $$2x^2 + Kx + 8 = 0 \quad \text{and} \quad 3x^2 + 4x + 12 = 0$$ have both roots common.​

Correct option is A

Given:

The two quadratic equations are:

​$$2x^2 + Kx + 8 = 0$$​​

​$$3x^2 + 4x + 12 = 0$$​​

Concept Used:

For two quadratic equations to have both roots in common, the ratio of their corresponding coefficients must be equal:

​$$\frac{a}{p} = \frac{b}{q} = \frac{c}{r}$$​​

where $$ax^2 + bx + c = 0$$​ and $$px^2 + qx + r = 0$$​ are the two quadratic equations.

Solution:

For the first equation 2x^2 + Kx + 8 = 0:

a = 2, b = K, c = 8

For the second equation 3x^2 + 4x + 12 = 0:

p = 3, q = 4, r = 12

Apply the condition for common roots:

$$\frac{2}{3} = \frac{K}{4} = \frac{8}{12}$$​​

We will first solve $$\frac{2}{3} = \frac{K}{4}$$​ to find K:

$$\frac{2}{3} = \frac{K}{4}$$​​

K = $$\left( \frac{2}{3} \right) \times 4 = \frac{8}{3}$$​​

Simplifying $$\frac{8}{12}$$​ gives $$\frac{2}{3}$$​, which is the same as $$\frac{2}{3}$$​.

This confirms that the condition holds true for all ratios.

The value of K is $$\frac{8}{3}$$​​