Question

Find the values ofk for which the given equation has real and equal roots:​

Answer 1

Answer:-

Answer:-k² - 3k = 0 k = 0.3

Explanation:-

Given (k+1)x² - 2(k − 1)x + 1 = 0 has equal -

roots

Discriminant= 0

(-2 (k - 1)) ² - 4 (k + 1) = 0 -

4 (k² - 2k + 1) -4k-4 = 0

k² - 3k = 0 k = 0.3

Answer 2

Answer:

The given equation is (k – 12)x2 + 2(k – 12)x + 2 = 0

Here, a = k – 12, b = 2(k – 12) and c = 2

Since, the given equation has two equal real roots

Then we must have b2 – 4ac = 0

⇒ {2(k – 12)}2 – 4(k – 12) × 2 = 0

⇒ 4(k – 12)2 – 8(k – 12) = 0

⇒ 4(k – 12) {k – 12 – 2} = 0

⇒ (k – 12)(k – 14) = 0

⇒ k – 12 = 0 or k – 14 = 0

⇒ k = 12 or k = 14