Question

find the value of k if given equation has real and equal roots K^2x^2-(k-1) x+4=0​

Answer 1

Answer:

k = (-1/3) or k = (1/5)

Step-by-step explanation:

If a Quadratic equation has real and equal roots then its Discriminant must be 0

Now,

Discriminant = b² - 4ac

when the Quadratic equation is ax² + bx + c = 0

Now, we are given,

k²x² - (k - 1)x + 4 = 0

ax² + bx + c = 0

where a = k², b = -(k - 1), and c = 4

If they have real and equal roots then,

b² - 4ac = 0

(-(k - 1))² - 4(k²)(4) = 0

we know that,

(a × b)^m = (a^m) × (b^m)

Similarly,

(-(k - 1))² = (-1 × (k - 1))²

= (-1)² × (k - 1)²

= 1 × (k - 1)²

= (k - 1)²

Now, back to our equation,

(-(k - 1))² - 4(k²)(4) = 0

(k - 1)² - 16k² = 0

Now, (a - b)² = a² - 2ab + b²

(k² - 2(k)(1) + 1²) - 16k² = 0

k² - 2k + 1 - 16k² = 0

k² - 16k² - 2k + 1 = 0

-15k² - 2k + 1 = 0 (Multiplying whole equation by (-1))

15k² + 2k - 1 = 0

ak² + bk + c = 0

where a = 15, b = 2, c = -1

By Splitting the Middle term method,

Sum = b = 2

Product = a × c = -15

So, the factors are 5 and -3

15k² + 5k - 3k - 1 = 0

5k(3k + 1) -1(3k + 1) = 0

(3k + 1)(5k - 1) = 0

Thus, either

(3k + 1) = 0 or (5k - 1) = 0

3k = -1 or 5k = 1

k = (-1/3) or k = (1/5)

So, k can be (-1/3) or (1/5)

Hope it helped and you understood it........All the best