Question

. Find the values of K for which the equation x² +2kx + 1=0 has real roots.
​

Answer 1

Answer:

for discriminant > 0, the quadratic equation has two real root

d = b^2 - 4ac = (2k)^2 - 4(1)(k(k - 1))

d = 4k^2 - 4k^2 + 4k

d = 4k

for k > 0, d > 0

Answer 2

Answer:

K is greater than +/- 1

Step-by-step explanation:

if a equation has real roots then

b^2 - 4ac is greater than 0( here b is coefficient of x, a is coefficient of x^2 , whereas c is constant term)

so putting the value

We get

(2k)^2 -4 is greater than 0

4k^2 is greater then 4

k^2 is greater then 1

so k is greter then +/- 1