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.
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Answered by
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
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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