please solve number 3
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Kx(x - 2) + 6 = 0
Kx² - 2kx + 6 = 0
Compare ax² + bx + c =0
a = k b = -2k c =6
Δ = b² - 4ac = (-2k)² - (4 x k x 6) = 4k² - 24k
So roots are real andequal
∴ Δ = 0
4k² - 24k = 0
4k(k - 6) = 0
k = 0 or 6
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