Question

1. For what value of K will be the following quadratic equations
(K + 1)x2 - 4 KX + 9 = 0 have real & equal roots ? solve the equations​

Answer 1

Answer:

(K+1)2 - 4Kx + 9 =0

2K + 2 - 4Kx + 9 = 0

2K - 4Kx + 11 = 0

2K(1 - 2x) = -11

1-2x = -11/2K

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Answer 2

Answer:

(k + 1)x² - 4kx + 9 = 0

First comapring the equation with general form that is, ax² + bx + c = 0

Here, a = k + 1 , b = - 4k , c = 9

The given, roots are real and equal.

Discriminant = 0

D = b² - 4ac = 0

➳ (-4 k)² - 4(k + 1) (9) = 0

➳ +16 k² - 36(k + 1) = 0

➳ 16k² - 36k - 36 = 0

[Now, taking 4 as a common we get]

➳ 4(4k² - 9k - 9) = 0

➳ 4k² - 9k - 9 = 0

[Now,by splitting middle term we get]

➳ 4k² - 12k + 3k - 9 = 0

➳ 4k(k - 3) + 3(k - 3) = 0

➳ (4k + 3) (k - 3) = 0

➳ k = -3/4 or 3

Therefore, k = 3 and k = -3/9.