Question

In each of the following, find the value of k for which the given value is a solution of the given equation:
(i) 7x²+kx-3=0,x=2/3
(ii) x²-x(a+b)+k=0,x=a
(iii) kx²+√2x-4=0,x=√2
(iv)x²+3ax+k=0,x=-a

Answer 1

Answer:

(I) 7(2/3)² + k(2/3) = 3

7(4/9) + 2k/3 = 3

28/9 + 2k/3 = 3

28 + 6k = 27

6k = 27 - 28

k = -1/6

Answer 2

(i) Given : 7x² + kx - 3 = 0

Since, x = 2/3 is a solution of a given equation, so it will satisfy the equation.

On putting x = ⅔ in 7x² + kx - 3 = 0

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 x² -x(a + b) + k = 0

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 kx² + √2x − 4 = 0

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  x² + 3ax + k  = 0

(-a)² + 3a(-a) + k = 0

a² -  3a² + k = 0

-2a² + k = 0

k = 2a²

Hence, the value of k is 2a².

HOPE THIS ANSWER WILL HELP YOU…

 

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