how to find the value of a Nd b
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(a - b) , a and (a + b) are the zeros of x³ -6x² + 11x -6 .
sum of zeros = - coefficient of x²/coefficient of x³
(a - b) + a + (a + b) = -(-6)/1 = 6
3a = 6 => a = 2
products of zeros = - constant/coefficient of x³
(2 - b)(2)(2 + b) = -(-6)/1 = 6
(4 - b²) = 3
4 - 3 = b² => b² = 1
b = ±1
hence ,
a = 2 and b = ±1
sum of zeros = - coefficient of x²/coefficient of x³
(a - b) + a + (a + b) = -(-6)/1 = 6
3a = 6 => a = 2
products of zeros = - constant/coefficient of x³
(2 - b)(2)(2 + b) = -(-6)/1 = 6
(4 - b²) = 3
4 - 3 = b² => b² = 1
b = ±1
hence ,
a = 2 and b = ±1
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