Q.If the equation \(3x^2 - 6x + p = 0\) has real and equal roots, then the value of \(p\) is – (a) \(\cfrac{5}{3}\) (b) -\(\cfrac{1}{3}\) (c) -3 (d) 3
Answer: D
The equation \(3x^2 - 6x + p = 0\) is compared to the standard form \(ax^2 + bx + c = 0\), which gives:
\(a = 3\), \(b = -6\), and \(c = p\).

Since the roots are real and equal,
∴ Discriminant \(= 0\).
Therefore, \(b^2 - 4ac = 0\).
That is, \((-6)^2 - 4 \times 3 \times p = 0\).
or, \(36 - 12p = 0\).
or, \(-12p = -36\).

\(∴ p = 3\).
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