Question

If one of the zeros of quadratic polynomial (k-1)x2+kx+1 is -3then find the other zero

Answer 1

Hey Mate here's the solution...


(k-1)x²+kx+1=0

(k-1)-3²+k(-3)+1= 0

(k-1)9-3k+1=0

9k-9-3k+1=0

6k=8

k=8/6=4/3

Again in Eq. putting value of k

(4/3-1)x²+4/3x+1 = 0

(1/3x²+4/3x+1=0) × 3

x²+4x+3 = 0

x²+3x+1x+3=0

x(x+3)+1(x+3)=0

(x+3)(x+1)=0

x+3=0 , x+1=0
x= -3, x= -1

So, other zero of eq. is -1.




Hope it will help you.....

Answer 2

Step-by-step explanation:

GIVEN:-)

→ One zeros of quadratic polynomial = -3.

→ Quadratic polynomial = ( k - 1 )x² + kx + 1.

Solution:-

→ P(x) = ( k -1 )x² + kx + 1 = 0.

→ p(-3) = ( k - 1 )(-3)² + k(-3) + 1 = 0.

=> ( k - 1 ) × 9 -3k + 1 = 0.

=> 9k - 9 -3k + 1 = 0.

=> 6k - 8 = 0.

=> 6k = 8. .......

$\large \boxed{=> k = \frac{8}{6} = \frac{4}{3} }$

Hence, the value of ‘k’ is founded .