Question

if one of the zero of the quadratic polynomial(k-1)x^2+kx+1 is -3, then the value of k is.

Answer 1

ANSWER:

• The Value of 'k' = 4/3.

GIVEN:

• P(x) = (k-1)x²+Kx+1
• One zero of the polynomial = (-3)

TO FIND:

• The value of 'k'.

SOLUTION:

P(x) = (k-1)x²+Kx+1

Putting x = (-3) because (-3) is the zero of the polynomial.

when we will put x = (-3) . we get the remainder 0.( because (-3) is the zero of the polynomial.)

P(-3) = (k-1)(-3)² +(-3)k +1

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

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

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

=> 6k -8 = 0

=> 6k = 8

=> k = 8/6

=> k = 4/3

The Value of 'k' = 4/3.

NOTE:

• By substituting zero in the polynomial we get the remainder= 0

Answer 2

$f(x) = (k - 1)x² + kx + 1$

$f(3) = 0$

$(k - 1) × (3)² + (-3)k + 1 = 0$

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

$9k - 9 - 3k + 1 = 0$

$6k - 8 = 0$

$6k = 8$

$k = \frac{8}{6}$

$k = \frac{4}{3}$