Question

If the set of all \( a \in \mathbb{R} \), for which the equation \( 2x^2 + (a - 5)x + 15 = 3a \) has no real root, is the interval \( (\alpha, \beta) \), and \( X = \{ x \in \mathbb{Z} : \alpha<x<\beta \} \), then \( \sum_{x \in X} x^2 \) is equal to:

• A. 2109
• B. 2129
• C. 2139
• D. 2119

Hint:

For polynomial inequalities, consider the discriminant condition carefully to determine valid intervals.

Answer 1

The Correct Option is C

Step 1: Identifying the condition for no real roots. The given equation is: \[ 2x^2 + (a - 5)x + 15 = 3a \] Rearranging, \[ 2x^2 + (a - 5)x + 15 - 3a = 0 \] For no real roots, the discriminant must be negative: \[ (a - 5)^2 - 8(15 - 3a)<0 \] Expanding, \[ a^2 + 25 - 10a - 120 + 24a<0 \] \[ a^2 + 14a - 95<0 \] \[ (a + 19)(a - 5)<0 \] This inequality holds true for: \[ -19<a<5 \]

Step 2: Finding integers in this interval. The integer values between -19 and 5 are: \[ \{-18, -17, \ldots, -1, 0, 1, \ldots, 4\} \]

Step 3: Summing the squares of the values. \[ \sum_{x \in X} x^2 = (1^2 + 2^2 + \cdots + 4^2) + (1^2 + 2^2 + \cdots + 18^2) \] Using the sum of squares formula: \[ \sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6} \] \[ = \frac{4 \times 5 \times 9}{6} + \frac{18 \times 19 \times 37}{6} \] \[ = 30 + 2109 = 2139 \]

Answer 2

Step 1: Write the given quadratic equation.
\[ 2x^2 + (a - 5)x + 15 = 3a \] Rearrange it to the standard quadratic form:
\[ 2x^2 + (a - 5)x + (15 - 3a) = 0 \] This is a quadratic in \( x \).

Step 2: Condition for no real roots.
A quadratic has no real roots if its discriminant is less than zero:
\[ D = b^2 - 4ac < 0 \] Here, \( a_1 = 2 \), \( b_1 = (a - 5) \), and \( c_1 = (15 - 3a) \).
So, \[ D = (a - 5)^2 - 4(2)(15 - 3a) \] Simplify: \[ D = a^2 - 10a + 25 - 8(15 - 3a) \] \[ D = a^2 - 10a + 25 - 120 + 24a \] \[ D = a^2 + 14a - 95 \] For no real roots: \[ a^2 + 14a - 95 < 0 \]

Step 3: Solve the inequality.
Find the roots of the quadratic equation:
\[ a^2 + 14a - 95 = 0 \] \[ a = \frac{-14 \pm \sqrt{14^2 - 4(1)(-95)}}{2} \] \[ a = \frac{-14 \pm \sqrt{196 + 380}}{2} = \frac{-14 \pm \sqrt{576}}{2} \] \[ a = \frac{-14 \pm 24}{2} \] \[ a_1 = -19, \quad a_2 = 5 \] So, the inequality \( a^2 + 14a - 95 < 0 \) holds for:
\[ -19 < a < 5 \] Hence, \( (\alpha, \beta) = (-19, 5) \).

Step 4: Define the set \( X \).
\[ X = \{ x \in \mathbb{Z} : \alpha < x < \beta \} \] \[ X = \{ -18, -17, -16, \dots, 3, 4 \} \] This is a sequence of integers from -18 to 4.

Step 5: Find the sum of squares of elements in X.
We need: \[ \sum_{x \in X} x^2 = (-18)^2 + (-17)^2 + \dots + 4^2 \] Use the formula for sum of squares: \[ \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6} \] Here we must handle negative and positive parts separately.

Sum of squares from 1 to 18: \[ \sum_{k=1}^{18} k^2 = \frac{18 \times 19 \times 37}{6} = 2109 \] Include 0? (not in range). Add squares from 1² + 2² + 3² + 4² for positives (since negatives are already included symmetrically except sign doesn’t matter). But our range is -18 to +4, so we include 1² + 2² + 3² + 4² again.

\[ \sum_{k=1}^{4} k^2 = 1 + 4 + 9 + 16 = 30 \] Thus total: \[ 2109 + 30 = 2139 \]

Final Answer:
\[ \boxed{2139} \]