Question

Find the value(s) of k for which each of the following quadratic equation has equal roots: 3kx² = 4(kx – 1)

Answer 1

3kx² = 4(kx – 1)
⇒ 3kx² = 4kx – 4
⇒ 3kx² – 4kx + 4 = 0
Here a = 3k, b = –4k, c = 4
∴ D = b² – 4ac
= (–4k)² – 4 x 3k x 4
= 16k² – 48k
∴ Roots are equal
∴ D = 0
⇒ 16k² – 48k = 0
⇒ k² – 3k = 0
⇒ k(k – 3) = 0
Either k = 0
or
k - 3 = 0
⇒ then k = 3

∴ x = `(-b ± sqrt("D"))/(2a) = (-b)/(2a)`  ...(∵ D = 0)

`(4k)/(2 xx 3k)`

= `(4 xx 3)/(2 xx 3 xx 3)`

= `(12)/(18)`

= `(2)/(3)`

∴ x = `(2)/(3), (2)/(3)`.