Question

if one of the zeroes of quadratic polynomial 5x^2-kx+2(k+1)is 1 .find the value of k​

Answer 1

Answer:

We know :

$\tt{\alpha + \beta = \frac{-b}{a}}$

One zero = 1

$\tt{\beta = \frac{k}{5} - 1}$

=》 $\tt{\beta = \frac{k-5}{5}}$ ...(1)

We even know :

$\tt{\alpha\beta = \frac{c}{a}}$

One zero = 1

=》 $\tt{\beta = \frac{2k+2}{5}}$ ...(2)

(1) = (2)

=》 k - 5 = 2k + 2

=》 k = 7

Hope it Helps ya! ❤

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 .