Question

If \(\alpha\) and \(\beta\) are two zeroes of a polynomial \(f(x) = px^2 - 2x + 3p\) and \(\alpha + \beta = \alpha\beta\), then value of p is :

• A. \(-\frac{2}{3}\)
• B. \(\frac{2}{3}\)
• C. \(\frac{1}{3}\)
• D. \(-\frac{1}{3}\)

Hint:

For any quadratic polynomial \(ax^2 + bx + c\), if the sum of roots equals the product of roots, then we can directly use the relation:
\[ -b = c \]
Here, with \(b = -2\) and \(c = 3p\):
\[ -(-2) = 3p \implies 2 = 3p \implies p = \frac{2}{3} \]
Using this direct shortcut bypasses the need to write out fractions and saves valuable time!

Answer 1

The Correct Option is B

Step 1: Write both Vieta relations in one line before substituting anything.
For $f(x) = px^2 - 2x + 3p$, matching with $ax^2 + bx + c$ gives $a = p$, $b = -2$, $c = 3p$. So:
\[ \alpha + \beta = \frac{2}{p}, \qquad \alpha\beta = \frac{3p}{p} = 3 \]

Step 2: Put the given condition into a single equation and clear the fraction in one move.
The condition is $\alpha + \beta = \alpha\beta$. Substitute both expressions at once:
\[ \frac{2}{p} = 3 \]
Instead of solving this in two separate stages, cross multiply directly:
\[ 2 = 3p \]

Step 3: Solve the resulting linear equation.
\[ p = \frac{2}{3} \]

Step 4: Sanity check.
With $p = \frac{2}{3}$: sum of zeroes $= \frac{2}{p} = \frac{2}{2/3} = 3$, and product of zeroes $= 3$. Since sum equals product, the condition holds, confirming the value is correct.

Step 5: Final Answer.
The value of $p$ is $\frac{2}{3}$, so option (B) is correct. \[ \boxed{p = \frac{2}{3}} \]