Question

if one of the zeros of the quadratic polynomial (k-1)x²+kx+1 is -3 then the value of k will be?

Answer 1

$\huge \boxed{ \mathbb{ANSWER:}}$

$<b> <I> <H3>$

$<u> GIVEN:-) </u>$

→ One zeros of quadratic polynomial = -3.

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

$<marquee> <u> Solution:- </u> </marquee>$

→ 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.

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

$<marquee> <u> ✔✔Hence, the value of ‘k’ is founded . <HR>$

$</H2> </b> </u> </I> </marquee>$

$\huge \boxed{ \mathbb{THANKS}}$

$\huge \bf{ \# \mathbb{B}e \mathbb{B}rainly.}$

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 .