Question

2x^2+kx-4=0 has real roots..find value of k

Answer 1

Since the equation has real roots

b^2 - 4ac = 0 ...... (i)

In the equation 2x^2 + kx - 4

a = 2, b = k and c = -4

Substituting the values of a,b and c in (i)

k^2 - 4*2*(-4) =0

k = √-32

k = +- 4√2

Answer 2

2x²+kx-4=0

D≥0

k² -4×2×-4  ≥ 0

k²+32 ≥ 0

k² ≥ -32

k ≥ √-32

k ≥  i√32