Question

If \( \alpha, \beta \) are roots of equation \( x^{2}+6 x+\lambda=0 \) and \( 3 \alpha+2 \beta=-20 \), then \( \lambda \) is equal to 

(A) 16 

(C) \( -16 \) 

(B) \( -8 \) 

(D) 8

Answer 1

Correct option is (c) -16

Sum of roots \(=\alpha + \beta = -6\)  ....(1)

Given that  \(3\alpha + 2\beta = -20\)    ....(2)

Now, by equation (2) - 2 x equation (1), we get

\(3\alpha + 2\beta -(2\alpha + 2\beta )= -20-(-12)\)

⇒ \(\alpha = -20 + 12 = -8\)

\(\therefore \beta = -6-\alpha = -6-(-8) = -6 + 8 = 2\)

\(\therefore\) Product of roots \(= \frac \lambda1= \lambda\)

\(\therefore \lambda = \alpha \beta = -8 \times 2 = -16\)

Comments

The way the problem is stated, x^2 + 6x + lambda ......
alpha plus beta equals +6, not -6

---

α+β = -b/a
= -(6)/1
= -6