Let the mean and the variance of seven observations 2, 4, , 8, , 12, 14, < , be 8 and 16 respectively. Then the quadratic equation whose roots are 3 +2 and 2 +1 is :
- x2-35x+306=0
- x2-41x+420=0
- x2-45x+506=0
- x2-37x+342=0
Answer
(2) x2-41x+420=0
Solution
Mean = 8, so 2 + 4 + + 8 + + 12 + 147 = 8
40 + + = 56 + = 16
Variance = 16, so xi27 - 82 = 16 xi2 = 560
4 + 16 + 64 + 144 + 196 + 2 + 2 = 560 2 + 2 = 136
2 = ( + )2 - ( 2 + 2 ) = 256 - 136 = 120 = 60
So α, β are the roots of t2 - 16t + 60 = 0, i.e. t = 6, 10, and < , = 6, = 10.
The required roots are 3 + 2 = 20 and 2 + 1 = 21, with sum = 41 and product = 420.
Required equation: x2 - 41x + 420 = 0
Answer: (2)