Key Concepts and Formulas

• Discriminant of a quadratic equation $Ax^2 + Bx + C = 0$: $D = B^2 - 4AC$. The equation has no real roots if $D < 0$.
• Sum of first $n$ squares: $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$
• Properties of squares: $(-k)^2 = k^2$

Step-by-Step Solution

Step 1: Rewrite the Equation in Standard Form

We need to rewrite the given equation $2x^2 + (a-5)x + 15 = 3a$ in the standard quadratic form $Ax^2 + Bx + C = 0$. This allows us to easily identify the coefficients A, B, and C. $2x^2 + (a-5)x + 15 - 3a = 0$ $2x^2 + (a-5)x + (15 - 3a) = 0$ Now we have:

• $A = 2$
• $B = a - 5$
• $C = 15 - 3a$

Step 2: Apply the Discriminant Condition for No Real Roots

Since the equation has no real roots, the discriminant must be less than zero, i.e., $D < 0$. We substitute the coefficients into the discriminant formula $D = B^2 - 4AC$. $(a-5)^2 - 4(2)(15-3a) < 0$ Now, we expand and simplify the inequality: $(a^2 - 10a + 25) - 8(15-3a) < 0$ $a^2 - 10a + 25 - 120 + 24a < 0$ $a^2 + 14a - 95 < 0$

Step 3: Solve the Quadratic Inequality to Find the Interval $(\alpha, \beta)$

To solve the inequality $a^2 + 14a - 95 < 0$, we first find the roots of the corresponding quadratic equation $a^2 + 14a - 95 = 0$. We factor the quadratic expression by finding two numbers that multiply to $-95$ and add up to $14$. These numbers are $19$ and $-5$. $(a+19)(a-5) = 0$ The roots are $a = -19$ and $a = 5$. Since the coefficient of $a^2$ is positive, the parabola opens upwards. Thus, the inequality $a^2 + 14a - 95 < 0$ is satisfied when $a$ is between the roots. $-19 < a < 5$ Comparing this to the interval $(\alpha, \beta)$, we get $\alpha = -19$ and $\beta = 5$.

Step 4: Identify the Set $X$ of Integers

The set $X$ is defined as all integers $x$ such that $\alpha < x < \beta$. Substituting the values of $\alpha$ and $\beta$: $X = \{x \in \mathbb{Z} \mid -19 < x < 5\}$ The integers strictly greater than $-19$ and strictly less than $5$ are: $X = \{-18, -17, -16, ..., -1, 0, 1, 2, 3, 4\}$

Step 5: Calculate the Sum of Squares for $x \in X$

We need to calculate $\sum_{x \in X} x^2$, which means summing the squares of all integers in $X$: $\sum_{x \in X} x^2 = (-18)^2 + (-17)^2 + \dots + (-1)^2 + 0^2 + 1^2 + 2^2 + 3^2 + 4^2$ Since $(-k)^2 = k^2$, we can rewrite the sum as: $\sum_{x \in X} x^2 = 18^2 + 17^2 + \dots + 1^2 + 0 + 1^2 + 2^2 + 3^2 + 4^2$ $\sum_{x \in X} x^2 = \sum_{k=1}^{18} k^2 + \sum_{k=1}^{4} k^2$ Using the formula for the sum of the first $n$ squares: $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$

For the first part ($n=18$): $\sum_{k=1}^{18} k^2 = \frac{18(18+1)(2(18)+1)}{6} = \frac{18(19)(37)}{6} = 3(19)(37) = 2109$

For the second part ($n=4$): $\sum_{k=1}^{4} k^2 = \frac{4(4+1)(2(4)+1)}{6} = \frac{4(5)(9)}{6} = \frac{180}{6} = 30$

Finally, we add the two sums: $\sum_{x \in X} x^2 = 2109 + 30 = 2139$

Common Mistakes & Tips

• Remember to consider the case when the discriminant is less than zero for the quadratic to have no real roots.
• Be careful with the signs when expanding and simplifying the quadratic inequality.
• Don't forget to include $0$ in the summation if it falls within the interval.

Summary

We started by rewriting the given quadratic equation in standard form and using the discriminant condition for no real roots to obtain a quadratic inequality in terms of $a$. Solving this inequality gave us the interval $(\alpha, \beta)$ for $a$. Then we identified the set of integers $X$ within this interval and calculated the sum of the squares of these integers using the sum of squares formula. The final sum is 2139.

Final Answer The final answer is $\boxed{2139}$, which corresponds to option (A).