Question

if alpha and beta are the zeroes of the quadratic polynomial 6x^2-(k+1)-7x such that one zero is11/6 more than the other,find the value of k
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Answer 1

Step-by-step explanation:

let the first zero be m

then the secone zero be (m+ 11/66)

eq. is 6x²-(k+1)-7x

6x²-7x-(k+1)

sum of root = -b/a

m+m+11/16)=-(-7/6)

2m+11/6= 7/6

2m= 7/6-11/6

2m= -4/6

m= -2/6

m= -1/3

second root = -1/3+11/6

= 9/6

= 3/2

product of root = c/a

-1/3×3/2= -(k+1)/6

-1/2= -(k+1)/6

1= (k+1)/3

3= k+1

2= k

for checking

(x+1/3)(x-3/2)=0

x²-3x/2+x/3- 1/2=0

x²-7x/6-1/2=0

on mutiply both side by 6

6(x²-7x/6-1/2)=0×6

6x²-7x-3 = 0

6x²-7x-(k+1) here k=2

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