Question

please solve this and get points​

Answer 1

Answer:

The given equation is

(k-3)x

$(k - 3)x { }^{2} + 4(k - 3)x + 4 = 0 \\ a = (k - 3) \: b = 4(k - 3) \: c = 4 \\ b { \: }^{2} - 4ac = (4k - 3) ^{2} - 4 \times (k - 3) \times 4 \\ = 4k {}^{2} \: - 24k + 9 - 4k + 12 + 4 \\ = 4k {}^{2} - 24k - 4k + 9 + 12 + 4 \\ = 4k {}^{2} - 28k + 25 = 0$