If the quadratic equation \((x + 1)(x + 2) = k(3x + 7)\) has equal roots, find the possible values of the constant k.
\((x + 1)(x + 2) = k(3x + 7)\)
\(x^{2} + 3x + 2 = 3kx + 7k \)
\(x^{2} + 3x - 3kx + 2 - 7k = 0\)
\(x^{2} + (3 - 3k)x + 2 - 7k = 0\)
For equal roots, \(b^{2} - 4ac = 0, a \neq 0\)
\(\therefore (3 - 3k)^{2} - 4(1)(2 - 7k) = 0\)
\(9 - 18k + 9k^{2} - 8 + 28k = 0\)
\(9k^{2} + 10k + 1 = 0 \)
\(9k(k + 1) + 1(k + 1) = 0 \implies (9k + 1)(k + 1) = 0\)
\(\implies k = -\frac{1}{9} ; k = -1\)
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