Question

Find the value of k if one of the zeroes of the quadratic polynomial

(k-1)x2 + k x + 1 is -3.​

Answer 1

SOLUTION

TO DETERMINE

The value of k if one of the zeroes of the quadratic polynomial (k-1)x² + kx + 1 is - 3.

EVALUATION

Here the given polynomial is

(k-1)x² + kx + 1

Now it is given that one of its roots is - 3

So by the given condition

$\sf{(k - 1) {( - 3)}^{2} + k \times ( - 3) + 1 = 0}$

$\sf{ \implies \: 9(k - 1) - 3 k + 1 = 0}$

$\sf{ \implies \: 9k - 9 - 3 k + 1 = 0}$

$\sf{ \implies \: 6k - 8 = 0}$

$\sf{ \implies \: 6k = 8 }$

$\displaystyle \sf{ \implies \: k = \frac{8}{6} }$

$\displaystyle \sf{ \implies \: k = \frac{4}{3} }$

FINAL ANSWER

$\displaystyle \sf{ \: k = \frac{4}{3} }$

━━━━━━━━━━━━━━━━

Learn more from Brainly :-

1. Write the degree of the polynomial :

4z3 – 3z5 + 2z4 + z + 1

https://brainly.in/question/7735375

2. Find the degree of 2020?

https://brainly.in/question/25939171

Answer 2

Given   : one of the zeroes of the quadratic polynomial ( k-1)x² + k x + 1 is -3.​

To Find : Value of k

Solution:

one of the zeroes of the quadratic polynomial ( k-1)x² + k x + 1 is -3.​

Let say P(x) = ( k-1)x² + k x + 1

-3 is one of the zero

=> P(-3) = 0

=>   ( k-1)(-3)² + k (-3) + 1  = 0

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

=> 6k = 8

=> k = 8/6

=> k = 4/3

Value of k is 4/3

Learn More:

if 2 + root 3 and 2 minus root 3 are the two zeros of the polynomial 2 ...

brainly.in/question/8973942

if the product of two roots of the equation 4x^4-24x^3+31x^2+6x-8=0 ...

brainly.in/question/18325992