Question

If one root of the equation $2x^2 - 3x + k = 0$ be reciprocal of the other, then the value of $k$ is

• A. $\dfrac{3}{2}$
• B. $-\dfrac{3}{2}$
• C. 3
• D. 2

Hint:

If one root of a quadratic equation is the reciprocal of the other, then the product of the roots is always 1. Use $\alpha\beta = \dfrac{c}{a}$ immediately.

Answer 1

The Correct Option is D

Step 1: Understand the condition on the roots.}
Let the two roots of the quadratic equation be $\alpha$ and $\beta$. The question says that one root is the reciprocal of the other. This means: \[ \beta = \frac{1}{\alpha} \] Therefore, the product of the roots becomes: \[ \alpha \beta = \alpha \cdot \frac{1}{\alpha} = 1 \] So, for this quadratic equation, the product of the roots must be equal to 1.

Step 2: Use the product of roots formula.}
For a quadratic equation of the form: \[ ax^2 + bx + c = 0 \] the product of the roots is: \[ \alpha \beta = \frac{c}{a} \] Here, \[ a = 2, \qquad c = k \] So, \[ \alpha \beta = \frac{k}{2} \] But from the given condition, we already know that: \[ \alpha \beta = 1 \] Hence, \[ \frac{k}{2} = 1 \]

Step 3: Find the value of $k$.}
Multiplying both sides by 2, we get: \[ k = 2 \] Thus, the required value of $k$ is: \[ 2 \]

Step 4: Compare with the options.}

• (A) $\dfrac{3{2}$:} Incorrect. This does not make the product of roots equal to 1.
• (B) $-\dfrac{3{2}$:} Incorrect. This also does not satisfy the reciprocal condition.
• (C) 3: Incorrect. This would give product of roots $\dfrac{3}{2}$.
• (D) 2: Correct. This gives product of roots $\dfrac{2}{2} = 1$.

Therefore, the correct value of $k$ is 2.
Final Answer:} $2$.