Question

If the zeroes of the cubic polynomial x3 – 6x2 + 3x + 10 are of the form a,a + b and a + 2b for some real numbers a and b, find the values of a and b . *​

Answer 1

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Answer 2

Step-by-step explanation:

clearly x=2 is a solution of the eqn

factorising the eqn we get

X3 -2x2 -4x2 + 8x - 5x + 10 = 0

x2(x-2) - 4x(x-2) - 5(x-2) = 0

(x-2)(x2 -4x - 5) = 0

this gives

x-2 = 0, x= 2

x2 - 4x - 5 = 0

x2 - 5x + x - 5 = 0

x(x-5) + 1(x-5) = 0

(x-5)(x+1) = 0

x = -1, 5

so roots are -1, 2 and 5

case 1

a = -1 and a + b = 2

which gives b = 2 - a = 3

a + 2b = 5 which is satisfied with the values of a= -1 and b = 3.

hence a= -1, b =3 is a solution

case 2

a = -1 and a+b = 5

this gives b = 6

but a+2b = 2 is not satisfied with the value of a= -1 and b = 6, so is not a solution

case 3

a= 2 and a+b= -1

this gives b = -3

but a+2b = 5 is not satisfied with these vues, so is not a solution

case 4

a = 2 and a+b = 5

this gives b = 3

but a+2b = -1 is not satisfied hence is not a solution

case 5

a = 5 and a+b = 2

this gives b = -3

a+2b = -1 is satisfied with these values hence

a= 5 and b= -3 is also a solution

case 6

a= 5 and a+b = -1

this gives b= -6

a+2b = 2 is not satisfied with these values hence not a solution

hence the solutions in real numbers are

1. a= -1, b = 3

2. a = 5, b = -3