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
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