Question

If one of the zeroes of the quadratic polynomial (k-1)x2 +kx +1 is -3, then find the value of k.

Answer 1

Answer:

∴ k = 10/3

Step-by-step explanation:

Let α and β be the two zeroes of the polynomial (k-1)x²+kx+1.

Here, α = -3

Sum of the zeroes is,

α+β = -k

or, -3+β = -k

∴ β = 3-k ------- (1)

Again,

Product of the roots is,

αβ = 1

or, -3×β = 1

∴ β = -1/3 ------- (2)

From equation (1) and (2),

3-K = -1/3

or, 3+1/3 = k

∴ k = 10/3

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 .