Question

solve for k
$(k - 1)x { }^{2} + kx + 1$
if one zero is -3 of this quadratic polynomial

Answer 1

Answer:

Step-by-step explanation:

you have to find a number by writing this equation =0

i.e.  5xsquare-kx1+2(k+1) 

by putting the value of x & k we get after solving this equation should be =0

you must have got the value of x in your question

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 .