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