Question

Please answer this!
3. 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
(ii) kor2 + 12x – 4 = 0, x = 12
(iv) x² + 3ax + k = 0, x = -a​

Answer 1

Answer:

i.

$7 {x}^{2} + kx - 3 = 0 \: \: \\ x = \frac{2}{3} \\ 7( { \frac{2}{3} })^{2} + \frac{2k}{3} - 3 = 0 \\ \frac{28}{9} + \frac{2k}{3} = 3 \\ \frac{28 + 6k}{9} = 3 \\ 28 + 6k = 27 \\ 6k = 27 - 28 \\ 6k = - 1 \\ k = - \frac{1}{6}$

ii.

Answer 2

Answer:

(1) 14-3=2/3k

k=36

(2)

2a-a×a-ab+k=0

k=2a-a×a+ab