Question

Y= 3 is a root of 7y^2-(k+1)y+3=0 if K​

Answer 1

Answer:

$7 {y}^{2} - (k + 1)y + 3 = 0$

$7 \times ({3})^{2} - (k + 1) \times 3+ 3 = 0$

$7 \times 9 - 3k - 3 + 3 = 0$

$63 - 3k = 0$

$- 3k = - 63$

$k = \frac{63}{3}$

$k = 12$

The value of k is 12.