Question

Let \(\alpha\) and \(\beta\) be the two distinct roots of the equation of 2x²-6x+k=0, such that (\(\alpha+\beta\)) and \(\alpha\beta\) are the distinct roots of the equation x²+px+p=0, then, the value of 8(k-p) ?

• A. 1
• B. 4
• C. 6
• D. 3

Answer 1

The Correct Option is C

Given: α and β are the distinct roots of the quadratic equation: 

\[ 2x^2 - 6x + k = 0 \]

From the standard form of a quadratic equation:

- Product of roots \( \alpha\beta = \frac{k}{2} \) - Sum of roots \( \alpha + \beta = \frac{-(-6)}{2} = 3 \)

Now, (α + β) and αβ are roots of another equation:

\[ x^2 + px + p = 0 \]

Let the roots of this new equation be \( r_1 = \alpha + \beta = 3 \) and \( r_2 = \alpha\beta = \frac{k}{2} \)

Using sum of roots:

\[ r_1 + r_2 = 3 + \frac{k}{2} = -p \tag{1} \]

Using product of roots:

\[ r_1 \cdot r_2 = 3 \cdot \frac{k}{2} = p \tag{2} \]

Now substitute \( p = \frac{3k}{2} \) from (2) into equation (1):

\[ 3 + \frac{k}{2} = -\frac{3k}{2} \]

Multiply both sides by 2 to eliminate denominator:

\[ 6 + k = -3k \Rightarrow 6 = -4k \Rightarrow k = -\frac{3}{2} \]

Now substitute \( k = -\frac{3}{2} \) into (2) to get p:

\[ p = \frac{3k}{2} = \frac{3}{2} \cdot \left(-\frac{3}{2}\right) = -\frac{9}{4} \]

Now compute:

\[ 8(k - p) = 8\left(-\frac{3}{2} - (-\frac{9}{4})\right) = 8\left(-\frac{3}{2} + \frac{9}{4}\right) \]

Convert to common denominator:

\[ -\frac{6}{4} + \frac{9}{4} = \frac{3}{4} \Rightarrow 8 \cdot \frac{3}{4} = \boxed{6} \]

Final Answer: 6 (Option C)