Question

20. For what value of k, the quadratic equation ky² - 4y + 1 = 0 is a perfect square ?​

Answer 1

Answer: 4

Step-by-step explanation: we use the formula b²-4ac

so ∴ b=-4 a=k c=1

so, we substitute the values in the equation and we get

(-4)²-4×k×1

=16-4k=0

16=4k

k=16/4

k=4

Answer 2

Answer:

2

Step-by-step explanation:

ky²-4y+1 = 0

As we know that (a+b)² = a² +b² + 2ab

Therefore

By trial and error method K = 4

Because (2y)² - 2(1)(2) y + 1²

= (2y -1 )

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