Question

For what value of k are the roots of the quadratic equation x² - (3k-1)x+2k² +2k-11=0 are equal ​

Answer 1

Answer:

k = 9, 5

Step-by-step explanation:

for equal roots, D = 0

I.e. b² - 4ac = 0

so,(3k - 1)² - 4 ×1 × (2k² +2k - 11) = 0

9k² - 6k + 1 - 8k² - 8k + 44 =0

or, 9k²- 8k²- 6k- 8k+ 1+ 44 = 0

or, k² - 14k +45 = 0

or, k² - 9k - 5k + 45 = 0

or, k(k - 9) -5(k - 9) = 0

or, (k - 9)(k - 5) = 0

if k - 9 = 0 then k = 9

if k - 5 = 0 then k = 5

Answer 2

Answer:

k = 9 or 5

Step-by-step explanation:

Given Quadratic equation → x² - (3k-1)x + [2k²+2k-11] = 0

For roots to be equal, the value of Discriminant should be equal to zero.

D = 0

⇒ b² - 4ac = 0

⇒ b² = 4ac

⇒ (3k – 1)² = 4 (1) (2k² + 2k - 11)

⇒ (3k – 1)² = 4 (2k² + 2k – 11)

⇒ (3k)² + (1)² – 2(3k)(1) = 8k² + 8k – 44

⇒ 9k² + 1 – 6k = 8k² + 8k – 44

⇒ 9k² – 8k² + 1 + 44 = 8k + 6k

⇒ k² + 45 = 14k

⇒ k² - 14k + 45 = 0

⇒ k² – 9k – 5k + 45 = 0

⇒ k (k – 9) -5 (k – 9) = 0

⇒ (k – 9) (k – 5) = 0

∴ k = 9 or 5