Question

Find the values of k for which each of the following quadratic equation has equal roots: x² – 2kx + 7k – 12 = 0 Also, find the roots for those values of k in each case.

Answer 1

x² – 2kx + 7k – 12 = 0
Here a = 1, b = -2k., c = 7k - 12
∴ D = b² - 4ac
= (-2k)² - 4 x 1 x (7k - 12)
= 4k² - 4(7k - 12)
= 4k² - 28k + 48
∵ Roots are equal
∴ D = 0
⇒ 4k² - 28k + 48 = 0
⇒ k² - 7k + 12 = 0
⇒ k² - 3k - 4k + 12 = 0
⇒ k(k - 3) -4(k - 3) = 0
⇒ (k - 3)(k - 4) = 0
Either k - 3 = 0,
then k = 3
or
k - 4 = 0,
then k = 4
(a) If k = 3, then

x = `(-b ± sqrt("D"))/(2a)`

= `(4k ± sqrt(10))/(2 xx 1)`

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

= `(12)/(2)`
= 6
x = 6, 6
(b) If k = 4, then

x = `(-b ± sqrt("D"))/(2a)`

= `(-(-2 xx 4) ± sqrt(0))/(2 xx 1)`

= `(±8)/(2)`
= 4
∴ x = 4, 4.