Question

Find the value(s) of k for which each of the following quadratic equation has equal roots: (k + 4)x² + (k + 1)x + 1 =0 Also, find the roots for that value (s) of k in each case.

Answer 1

(k + 4)x² + (k + 1)x + 1 =0
Here a = k + 4, b = k + 1, c = 1
∴ D = b²  4ac
= (k + 1)² – 4 x (k + 4) x 1
= k² + 2k + 1 – 4k – 16
= k² - 2k - 15
∵ Root are equal
∴ k² – 2k – 15 = 0
⇒ k² – 5k + 3k – 15 = 0
⇒ k(k – 5) + 3(k – 5) = 0
⇒ (k – 5)(k + 3) = 0
Either k – 5 = 0, 
then k = 5
or
k + 3 = 0,
then k = –3
(a) When k = 5, then

x = `(-b ± sqrt("D"))/(2a) = (-b)/(2a)`

= `(-k - 1)/(2(k + 4)) = (-5 - 1)/(2(5 + 4)`

= `(-6)/(18) = (-1)/(3)`

∴ x = `(-1)/(3), (-1)/(3)`
(b) When k = –3, then

x = `(-b ± sqrt("D"))/(2a) = (-b)/(2a)`

= `(-k - 1)/(2(k + 4)) = ((-3) - 1)/(2(-3 + 4)`

= `(2)/(2 xx 1)` = 1
∴ x = 1, 1.