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

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

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Answer 1 · 2017-07-07T23:27:08

$\frac{a^{2}}{b + 1} = 3$ $a^2/(b+1) = 3$

Explanation:

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

$(x - x_{1}) (x - x_{2}) (x - x_{3})$ $(x-x_1)(x-x_2)(x-x_3)$

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

$(x - x_{1}) (x - (x_{1} + 1)) (x - (x_{1} + 2))$ $(x-x_1)(x-(x_1+1))(x-(x_1+2))$

Distribute the minus signs:

$(x - x_{1}) (x - x_{1} - 1) (x - x_{1} - 2)$ $(x-x_1)(x-x_1-1)(x-x_1-2)$

Here is the results of the multiplication:

$x^{3} + (- 3 x_{1} - 3) x^{2} + (3 x_{1}^{2} + 6 x_{1} + 2) x + (- x_{1}^{3} - 3 x_{1}^{2} - 2 x_{1})$ $x^3+(-3x_1-3)x^2+ (3x_1^2+6x_1+2)x+ (-x_1^3-3x_1^2-2x_1)$

Matching coefficients with $x^{2} + a x^{2} + b x + c$ $x^2+ax^2+bx+c$

$a = - 3 x_{1} - 3$ $a = -3x_1-3$

$b = 3 x_{1}^{2} + 6 x_{1} + 2$ $b = 3x_1^2+6x_1+2$

$c = - x_{1}^{3} - 3 x_{1}^{2} - 2 x_{1}$ $c = -x_1^3-3x_1^2-2x_1$

Compute $b + 1$ $b+1$ :

$b + 1 = 3 x_{1}^{2} + 6 x_{1} + 3$ $b+1 = 3x_1^2+6x_1+3$

$b + 1 = 3 (x_{1}^{2} + 2 x_{1} + 1) [1]$ $b+1 = 3(x_1^2+2x_1+1)" [1]"$

Compute $a^{2}$ $a^2$ :

$a^{2} = {(- 3 x_{1}^{2} - 3)}^{2}$ $a^2 = (-3x_1^2-3)^2$

$a^{2} = {(- 3 (x_{1} + 1))}^{2}$ $a^2= (-3(x_1+1))^2$

$a^{2} = 9 (x_{1}^{2} + 2 x_{1} + 1) [2]$ $a^2 = 9(x_1^2+2x_1+1)" [2]"$

Divide equation [2] by equation [1]:

$\frac{a^{2}}{b + 1} = \frac{9 (x_{1}^{2} + 2 x_{1} + 1)}{3 (x_{1}^{2} + 2 x_{1} + 1)}$ $a^2/(b+1) = (9(x_1^2+2x_1+1))/(3(x_1^2+2x_1+1))$

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

$\frac{a^{2}}{b + 1} = \frac{9}{3}$ $a^2/(b+1) = 9/3$

$\frac{a^{2}}{b + 1} = 3$ $a^2/(b+1) = 3$

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

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Explanation:

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The roots of the equations x^3+ax^2+bx+c=0x3+ax2+bx+c=0x^3+ax^2+bx+c=0 are three consecutive integers. Find all possible values of a^2/(b+1)a2b+1a^2/(b+1)?

I have no idea how to get there. Please include steps and explanations!

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