JAMB 1985 · UME · Q13

Quadratics & their graphs

Find the values of $p$ for which the equation $x^2 - (p - 2)x + 2p + 1 = 0$ has equal roots.

Worked solution (try it first)

Equal roots means $b^2 = 4ac$ .
Here $a = 1$ , $b = -(p - 2)$ and $c = 2p + 1$ .
So $(p - 2)^2 = 4(2p + 1)$ , which is $p^2 - 4p + 4 = 8p + 4$ .
Simplify: $p^2 - 12p = 0$ , so $p(p - 12) = 0$ .
So $p = 0$ or $p = 12$ , option A.

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