. If one of the zeroes of the quadratic polynomial (k-1) x2 +k x +1 is -3 then what is the value of k?

k = 1¹/₃

Step-by-step explanation:

Comparing (k-1)x² + kx + 1 with ax + bx + c where α, β be the zeroes of the quadratic equation, then

α + β = -b/a = -k/(k - 1) and αβ = c/a = 1/(k - 1)

Since one of the zeros is -3, β = -3

So,

α + β = -k/(k - 1)

α + (-3) = -k/(k - 1)

α - 3 = -k/(k - 1)  (1)

and

αβ = 1/(k - 1)

-3α = 1/(k - 1)    (2)

From (1), α = 3 - k/(k - 1)  (3)

Substituting equation (3) into (2), we have

-3α = 1/(k - 1)

-3[3 - k/(k - 1)] = 1/(k - 1)

-9 + 3k/(k - 1) = 1/(k - 1)

-9 = 1/(k - 1) - 3k/(k - 1)

-9 = (1 - 3k)/(k - 1)

cross-multiplying, we have

-9(k - 1) = 1 - 3k

expanding the brackets, we have

-9k + 9 = 1 - 3k

collecting like terms, we have

-9k + 3k = 1 - 9

-6k = -8

dividing through by -6, we have

k = -8/-6

k = 4/3

k = 1¹/₃

step-by-step explanation:

use pothagarean therom

a^2+b^2=c^2 so first ac

acm is a right triangle because c= 90°

and am (the hypotenuse or c) =16+9 =25

but there is not enough info we need another angle or side length

answer: i believe the answer is e

step-by-step explanation: