WAEC 2009 · Paper 2 · Q3

Polynomials & quadratic roots

(a)
If the quadratic equation $(x + 1)(x + 2) = k(3x + 7)$ has equal roots, find the possible values of the constant $k$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

Expand both sides: $x^2 + 3x + 2 = 3kx + 7k$ .
Bring everything to one side: $x^2 + (3 - 3k)x + (2 - 7k) = 0$ .
Equal roots need $b^2 = 4ac$ : $(3 - 3k)^2 = 4(2 - 7k)$ .
Expand: $9 - 18k + 9k^2 = 8 - 28k$ , so $9k^2 + 10k + 1 = 0$ .
Factorise: $(9k + 1)(k + 1) = 0$ .
So $k = -\frac19$ or $k = -1$ .

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