Question

x²-2x(1+3k)+7(3+2k)=0. find the value of k.

Answer 1

a = 1
b = 2(1 + 3k) = 6k + 2
c = 7 (3 + 2k) = 14 k + 21

$b {}^{2} - 4ac = 0$

$(6k + 2) {}^{2} - 4(1)(14k \: + \: 21) = 0$

$36k {}^{2} + 24k + \: 4 \: - 56k - 84 = 0$

$36k {}^{2} - 32k - 80 = 0$

Take 4 common

9k^2 - 8k - 20 = 0

9k^2 -18k + 10k - 20 = 0

9k (k - 2) + 10 (k - 2) = 0

(9k + 10)(k - 2) = 0

k = -10/9 and k = 2


Thanks