If the quadratic equation ((x + 1)(x + 2) = k(3x + 7)) has equal roots, find the possible values of the constant k.
Explanation
((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)