Question

Given, the equation x$^3$ - 75x + a = 0 has three integral roots $\alpha$, $\beta$ and $\gamma$ (where a $\in$ I) then $(\frac{|\alpha| + |\beta| + |\gamma|}{4})$ is equal to

• A. 2
• B. 4
• C. 5
• D. 6

Answer 1

Answer: (C)

5

Answer 2

Let the given cubic equation be $x^3 - 75x + a = 0$.
Let its three integral roots be $\alpha$, $\beta$, and $\gamma$.

Using Vieta's formulas for a cubic equation $Ax^3 + Bx^2 + Cx + D = 0$:

1. Sum of roots: $\alpha + \beta + \gamma = -B/A$
   For the given equation, $A=1$, $B=0$, $C=-75$, $D=a$.
   So, $\alpha + \beta + \gamma = -0/1 = 0$. (Equation 1)

2. Sum of products of roots taken two at a time: $\alpha\beta + \beta\gamma + \gamma\alpha = C/A$
   So, $\alpha\beta + \beta\gamma + \gamma\alpha = -75/1 = -75$. (Equation 2)

3. Product of roots: $\alpha\beta\gamma = -D/A$
   So, $\alpha\beta\gamma = -a/1 = -a$. (Equation 3)

From Equation 1, $\alpha + \beta + \gamma = 0$.
Since $\alpha, \beta, \gamma$ are integers, and their sum is zero, at least one root must be negative if there are positive roots, or vice versa. Also, no root can be zero, because if $\alpha=0$, then from Equation 3, $-a=0 \implies a=0$. The equation becomes $x^3 - 75x = 0 \implies x(x^2 - 75) = 0$. The roots would be $0, \sqrt{75}, -\sqrt{75}$, which are not all integers. Thus, all roots must be non-zero integers.

Substitute $\gamma = -(\alpha + \beta)$ from Equation 1 into Equation 2:
$\alpha\beta + \beta(-(\alpha + \beta)) + \alpha(-(\alpha + \beta)) = -75$
$\alpha\beta - \alpha\beta - \beta^2 - \alpha^2 - \alpha\beta = -75$
$-\alpha^2 - \beta^2 - \alpha\beta = -75$
$\alpha^2 + \beta^2 + \alpha\beta = 75$ (Equation 4)

We need to find integer solutions for $\alpha$ and $\beta$ that satisfy Equation 4.
We can rewrite Equation 4 as: $4\alpha^2 + 4\beta^2 + 4\alpha\beta = 300$
$(2\alpha + \beta)^2 + 3\beta^2 = 300$
Let $X = 2\alpha + \beta$. Then $X^2 + 3\beta^2 = 300$.
Since $\beta$ is an integer, $\beta^2$ must be a perfect square.
Also, $3\beta^2 \le 300 \implies \beta^2 \le 100 \implies -10 \le \beta \le 10$.
And $X^2 = 300 - 3\beta^2 = 3(100 - \beta^2)$. For $X^2$ to be a perfect square, $3(100 - \beta^2)$ must be a perfect square. This implies that $(100 - \beta^2)$ must be of the form $3k^2$ for some integer $k$.

Let's test integer values for $\beta$ within the range $[-10, 10]$ (excluding 0, as roots are non-zero):

• If $\beta = \pm 1$: $100 - (\pm 1)^2 = 99$, not $3k^2$.

• If $\beta = \pm 2$: $100 - (\pm 2)^2 = 96$, not $3k^2$.

• If $\beta = \pm 3$: $100 - (\pm 3)^2 = 91$, not $3k^2$.

• If $\beta = \pm 4$: $100 - (\pm 4)^2 = 84$, not $3k^2$.

• If $\beta = \pm 5$: $100 - (\pm 5)^2 = 75 = 3 \times 25 = 3 \times 5^2$. This is a valid case ($k=5$).
  If $\beta = 5$: $X^2 = 3(75) = 225 \implies X = \pm 15$.

  • Case A: $2\alpha + \beta = 15 \implies 2\alpha + 5 = 15 \implies 2\alpha = 10 \implies \alpha = 5$.
    Roots are $(\alpha, \beta, \gamma) = (5, 5, -(\alpha+\beta)) = (5, 5, -(5+5)) = (5, 5, -10)$.

  • Case B: $2\alpha + \beta = -15 \implies 2\alpha + 5 = -15 \implies 2\alpha = -20 \implies \alpha = -10$.
    Roots are $(\alpha, \beta, \gamma) = (-10, 5, -(\alpha+\beta)) = (-10, 5, -(-10+5)) = (-10, 5, 5)$. If $\beta = -5$: $X^2 = 3(75) = 225 \implies X = \pm 15$.

  • Case C: $2\alpha + \beta = 15 \implies 2\alpha - 5 = 15 \implies 2\alpha = 20 \implies \alpha = 10$.
    Roots are $(\alpha, \beta, \gamma) = (10, -5, -(\alpha+\beta)) = (10, -5, -(10-5)) = (10, -5, -5)$.

  • Case D: $2\alpha + \beta = -15 \implies 2\alpha - 5 = -15 \implies 2\alpha = -10 \implies \alpha = -5$.
    Roots are $(\alpha, \beta, \gamma) = (-5, -5, -(\alpha+\beta)) = (-5, -5, -(-5-5)) = (-5, -5, 10)$.

• If $\beta = \pm 6$: $100 - (\pm 6)^2 = 64$, not $3k^2$.

• If $\beta = \pm 7$: $100 - (\pm 7)^2 = 51$, not $3k^2$.

• If $\beta = \pm 8$: $100 - (\pm 8)^2 = 36$, not $3k^2$.

• If $\beta = \pm 9$: $100 - (\pm 9)^2 = 19$, not $3k^2$.

• If $\beta = \pm 10$: $100 - (\pm 10)^2 = 0 = 3 \times 0^2$. This is a valid case ($k=0$).
  If $\beta = 10$: $X^2 = 3(0) = 0 \implies X = 0$.

  • Case E: $2\alpha + \beta = 0 \implies 2\alpha + 10 = 0 \implies 2\alpha = -10 \implies \alpha = -5$.
    Roots are $(\alpha, \beta, \gamma) = (-5, 10, -(\alpha+\beta)) = (-5, 10, -(-5+10)) = (-5, 10, -5)$. If $\beta = -10$: $X^2 = 3(0) = 0 \implies X = 0$.

  • Case F: $2\alpha + \beta = 0 \implies 2\alpha - 10 = 0 \implies 2\alpha = 10 \implies \alpha = 5$.
    Roots are $(\alpha, \beta, \gamma) = (5, -10, -(\alpha+\beta)) = (5, -10, -(5-10)) = (5, -10, 5)$.

All these sets of roots $\{5, 5, -10\}$, $\{-10, 5, 5\}$, $\{10, -5, -5\}$, $\{-5, -5, 10\}$, $\{-5, 10, -5\}$, $\{5, -10, 5\}$ are permutations of the same set of integral roots: $\{5, 5, -10\}$.

Let's verify this set of roots:
For roots $(5, 5, -10)$:

1. Sum: $5 + 5 + (-10) = 0$. (Correct)
2. Sum of products: $(5)(5) + (5)(-10) + (-10)(5) = 25 - 50 - 50 = -75$. (Correct)
3. Product: $(5)(5)(-10) = -250$.
   From Equation 3, $\alpha\beta\gamma = -a$, so $-a = -250 \implies a = 250$. Since $a \in I$, this is a valid solution.

Now we need to calculate $(\frac{|\alpha| + |\beta| + |\gamma|}{4})$.
Using the roots $(5, 5, -10)$:
$|\alpha| = |5| = 5$
$|\beta| = |5| = 5$
$|\gamma| = |-10| = 10$

$(\frac{|\alpha| + |\beta| + |\gamma|}{4}) = (\frac{5 + 5 + 10}{4}) = (\frac{20}{4}) = 5$.