Question

The set of value(s) of k belongs to R for which, kx^2 –(k+1)x + 2k –1=0 has no real roots is:
(A) {1,-2}
(B) (-inf, -1/7) U (1,inf)
(C){-4}
(D) (2, 4)​

Answer 1

Answer:

For a quadratic equation to have real roots, discriminant must be greater than or equal to zero.

For the first equation,

k ² −4(1)(64)≥0 (∵discriminant=b²−4ac)

⇒ k² −256≥0

⇒ (k−16)(k+16)≥0

⇒ k≥16 and k≤−16

For the second equation,

64−4k≥0

⇒ k≤16

∴ the value of k that satisfies both the conditions is k=16.

∴Option D is correct.