Question

If the roots of the given equation $2x^2+3(\lambda-2)x+\lambda+4=0$ be equal in magnitude but opposite in sign, then $\lambda$ =

• A. 1
• B. 2
• C. 3
• D. 2/3

Answer 1

Answer: (B)

2

Answer 2

Let the roots of the quadratic equation $Ax^2+Bx+C=0$ be $\alpha$ and $\beta$. If the roots are equal in magnitude but opposite in sign, then $\beta = -\alpha$. The sum of these roots is $\alpha + (-\alpha) = 0$. According to Vieta's formulas, the sum of roots is also given by $-\frac{B}{A}$. Thus, $-\frac{B}{A} = 0$, which implies $B=0$. For the given equation $2x^2+3(\lambda-2)x+\lambda+4=0$, we have $A=2$, $B=3(\lambda-2)$, and $C=\lambda+4$. Setting $B=0$: $3(\lambda-2) = 0$ $\lambda-2 = 0$ $\lambda = 2$ We must also ensure that the roots are non-zero, which requires $C \neq 0$. For $\lambda=2$, $C = 2+4 = 6 \neq 0$. The equation becomes $2x^2+6=0$, leading to roots $\pm i\sqrt{3}$, which satisfy the condition.