Math, asked by KunalBaniwal, 8 months ago

find the value of K such that the polynomial x²-(K+6)x+22(K-1) has sum of its zeroes equal to half of the product.​

Answers

Answered by atahrv
2

Step-by-step explanation:

Given quadratic polynomial,

x² - (k+6)x + 22(k-1) = 0

By comparing it with ax²+bx+c = 0,we get

a = 1 , b = -(k+6) , c = 22(k-1)

Sum of zeroes = -b/a

= -[-(k+6)]/1

= k+6

Product of zeroes = c/a

= 22(k-1)/1

= 22(k-1)

As per problem,

Sum of zeroes = ½ × product of zeroes

k+6 = ½ × 22(k-1)

k+6 = 11k-11

17=10k

k=1.7

Answered by saikrishna1021
1

Answer: 4/3

Step-by-step explanation:alpha +beta = -b/a and alpha×beta=c/a

So -(-k+6)=1/2×22(k-1)

k-6=11k-11

k-11k=-11+6

-10k=-5

k=-5/-10 =1/2

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