WAEC 2012 · Paper 2 · Q2

(a)
For what values of $k$ are the roots of the equation $(k + 3)x^2 + (6 - 2k)x + k - 1 = 0$ real?
Show the answer
$k \le \frac32$ (with $k \ne -3$ for a quadratic)

Worked solution (try it first)

Read off the coefficients: $a = k + 3$ , $b = 6 - 2k$ , $c = k - 1$ .
Real roots need $b^2 - 4ac \ge 0$ : $(6 - 2k)^2 - 4(k + 3)(k - 1) \ge 0$ .
Expand the square: $(6 - 2k)^2 = 36 - 24k + 4k^2$ .
Expand the product: $4(k + 3)(k - 1) = 4(k^2 + 2k - 3)$
$= 4k^2 + 8k - 12$ .
Take one from the other: $36 - 24k + 4k^2 - 4k^2 - 8k + 12 \ge 0$ , so $48 - 32k \ge 0$ .
Add $32k$ to both sides: $48 \ge 32k$ .
Divide by 32: $k \le \frac32$ .
For the equation to be a quadratic, $a \ne 0$ , so also $k \ne -3$ .
The roots are real for $k \le \frac32$ ( $k \ne -3$ ).