Question

find the value of k for which the equation kx²+2x+1=0 has real and distict roots.

Answer 1

Given ::- kx²+2x+1

equal and real roots.

To Find ::- value of k

Solution ::-

Comparing the given equation with ,

ax² + bx + c =0

Here, a= k , b= 2 , c= 1

Now , as roots are real and equal,

b²-4ac = 0

(2)² - 4(k)(1) =0

4 - 4k = 0

- 4k = -4

k = -4/-4

k = 1

Therefore, the value of k is 1

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