If the set of all a ∈ ℝ, for which the equation 3x² + (a - 6)x + (5/4)a = 3 has no real root, is the interval (α, β) and X = {x ∈ ℤ; α < x < β}, then \(\sum_{x \in X} x^2\) is equal to:
Answer (2)
To find the values of 'a' such that the quadratic equation has no real roots, we calculate the discriminant and determine the interval (α, β). After identifying integer values in this interval, we compute their squares and find the sum. The method involves using properties of quadratic equations and integer arithmetic to complete the task.
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To solve this problem, we first need to determine the interval ( α , β ) for which the quadratic equation 3 x 2 + ( a − 6 ) x + 4 5 a = 3 has no real roots.
A quadratic equation a x 2 + b x + c = 0 has no real roots if its discriminant D = b 2 − 4 a c is less than zero.
Here, we have:
a = 3
b = a − 6
c = 4 5 a − 3
The discriminant D is given by: D = ( a − 6 ) 2 − 4 × 3 × ( 4 5 a − 3 ) = ( a − 6 ) 2 − 12 ( 4 5 a − 3 )
Simplify this: D = ( a 2 − 12 a + 36 ) − 15 a + 36 D = a 2 − 27 a + 72
Set the discriminant less than zero to find the interval: a 2 − 27 a + 72 < 0
To solve a 2 − 27 a + 72 < 0 , we can find the roots of the quadratic equation a 2 − 27 a + 72 = 0 :
Using the quadratic formula a = 2 a − b ± b 2 − 4 a c , we find: a = 2 × 1 27 ± ( − 27 ) 2 − 4 × 1 × 72 a = 2 27 ± 729 − 288 a = 2 27 ± 441 a = 2 27 ± 21
This gives the roots a 1 = 24 and a 2 = 3 .
The quadratic inequality a 2 − 27 a + 72 < 0 is satisfied in the interval ( 3 , 24 ) .
Thus, ( α , β ) = ( 3 , 24 ) .
Next, the set X = { x ∈ Z ; α < x < β } consists of integer values for x between 3 and 24, i.e., X = { 4 , 5 , 6 , … , 23 } .
Now calculate the sum ∑ x ∈ X x 2 :
The integers in X go from 4 to 23:
x = 4 ∑ 23 x 2 = 4 2 + 5 2 + 6 2 + … + 2 3 2
To compute this:
Calculate the square of each integer from 4 to 23.
Sum them all up.
This sum equals 4324.
Therefore, the final answer is: x ∈ X ∑ x 2 = 4324
