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(i) Given : 7x² + kx - 3 = 0
Since, x = 2/3 is a solution of given equation, so it will satisfy the equation.
7(⅔)² + k(2/3) −3 = 0
7(4/9)+ 2k/3 − 3 = 0
28/9 + 2k/3 - 3 = 0
2k/3 = 3 - 28/9
2k/3 = (27−28)/9
2k/3 = −1/9
k = - 1/9 × 3/2
k = - ⅙
Hence, the value of k is ⅙.
(ii) Given : x² -x(a + b) + k = 0
Since, x = a is a solution of given equation, so it will satisfy the equation.
On putting x = a in Given equation,
a² - a(a + b) + k = 0
a² - a² - ab + k = 0
-ab + k = 0
k = ab
Hence, the value of k is ab.
(iii) Given : kx² + √2x − 4 = 0
Since, x = √2 is a solution of given equation, so it will satisfy the equation.
On putting x = √2 in Given equation,
k(√2)² - √2 × √2 −4 = 0
2k + 2 - 4 = 0
2k - 2 = 0
2k = 2
k = 2/2
k = 1
Hence, the value of k is 1.
(iv) Given : x² + 3ax + k = 0
Since, x = a is a solution of given equation, so it will satisfy the equation.
On putting x = - a in Given equation,
(-a)² + 3a(-a) + k = 0
a² - 3a² + k = 0
-2a² + k = 0
k = 2a²
Hence, the value of k is 2a².
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