In each of the following, find the value of k for which the given value is a solution of the given equation:

(i)

(ii)

(iii)

(iv)

For the given value to be a solution, it should satisfy the quadratic equation

(i)

⇒ 7 × 4/9 + k × 2/3 – 3 = 0

⇒ 2k/3 = 3 – 28/9 = - 1/9

⇒ k = -1/6

(ii)

⇒ a² – a(a + b) + k = 0

⇒ a² –a² – ab + k = 0

⇒ k = ab

(iii)

⇒ k × 2 + √2 × √2 – 4 = 0

⇒ 2k = 2

⇒ k = 1

(iv)

⇒ a² – 3a × a + k = 0

⇒ k = 2a²