The roots of the equations #x^3+ax^2+bx+c=0# are three consecutive integers. Find all possible values of #a^2/(b+1)#?

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Answer 1 · 2017-07-07T23:27:08

#a^2/(b+1) = 3#

We know that the product of the factors for the 3 roots is equal to the polynomial:

#(x-x_1)(x-x_2)(x-x_3)#

Because we are told that the roots are consecutive integers, we can rewrite the above in terms of the lowest root:

#(x-x_1)(x-(x_1+1))(x-(x_1+2))#

Distribute the minus signs:

#(x-x_1)(x-x_1-1)(x-x_1-2)#

Here is the results of the multiplication:

#x^3+(-3x_1-3)x^2+ (3x_1^2+6x_1+2)x+ (-x_1^3-3x_1^2-2x_1)#

Matching coefficients with #x^2+ax^2+bx+c#

#a = -3x_1-3#

#b = 3x_1^2+6x_1+2#

#c = -x_1^3-3x_1^2-2x_1#

Compute #b+1#:

#b+1 = 3x_1^2+6x_1+3#

#b+1 = 3(x_1^2+2x_1+1)" [1]"#

Compute #a^2#:

#a^2 = (-3x_1^2-3)^2#

#a^2= (-3(x_1+1))^2#

#a^2 = 9(x_1^2+2x_1+1)" [2]"#

Divide equation [2] by equation [1]:

#a^2/(b+1) = (9(x_1^2+2x_1+1))/(3(x_1^2+2x_1+1))#

The quadratics cancel to show that the value of #a^2/(b+1)# has only 1 value, when the roots are consecutive integers:

#a^2/(b+1) = 9/3#

#a^2/(b+1) = 3#