Question

Find the value (s) of k for which each of the following quadratic equation has equal roots : (k – 4) x² + 2(k – 4) x + 4 = 0

Answer 1

(k – 4) x² + 2(k – 4) x + 4 = 0
Here a = k - 4, b = 2(k - 4), c = 4
D = b² - 4ac
= [2(k - 4)]² - 4 x (k - 4) x 4
= 4(k² + 16 - 8k) - 16(k - 4)
= 4(k² - 8k + 16) - 16(k - 4)
= 4[k² - 8k + 16 - 4k + 16]
= 4(k² - 12k + 32)
∵ Roots are equal
∴ D = 0
⇒ 4(k² - 12k + 32) = 0
⇒ k² - 12k + 32 = 0
⇒ k² - 8k - 4k + 32 = 0
⇒ k(k - 8) -4(k - 8) = 0
⇒ (k - 8)(k - 4) = 0
Either k - 8 = 0,
then k = 4
or
k - 4 = 0,
then k = 4
But k - 4 ≠ 0
k ≠ 4
k = 8.