Question

If one of the zeros of the quadric polynomial (k-1)x^{2} +kx+1 is -3, then find the value of k

Answer 1

HEYA USER ❤️

REFER TO THE ATTACHMENT ❣️❣️

Answer 2

$\huge \boxed{ \underline{ \underline{ \bf{Answer}}}}$

GIVEN :

$\sf{P(x) = \: {(k - 1)x}^{2} + kx + 1}$

One of the zeroes of P(x) = -3.

TO FIND :

Value of 'k'

SOLUTION :

$\sf{P(x) = \: {(k - 1)x}^{2} + kx + 1} = 0$

Putting the value of one of the zero.

$\sf{P( - 3) = \: {(k - 1)( - 3)}^{2} + k( - 3) + 1} = 0$

$\implies \sf {\: 9(k - 1) - 3k+ 1} = 0$

$\implies \sf {\: 9k -9 - 3k+ 1} = 0$

$\implies \sf{6k - 8 = 0}$

$\boxed{ \sf {\therefore \: \: k = \frac{8}{6} \: or \: \frac{4}{3} }}$