If the set of all a∈R, for which the equation 2x2+(a−5)x+15=3a has no real root, is the interval (α,β), and X={x∈Z:α<x<β}, then ∑x∈Xx2 is equal to:
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Step-by-Step Solution
Step 1: Rewrite the equation in standard quadratic form
First, we need to rearrange the given equation into the standard quadratic form Ax2+Bx+C=0. This involves moving all terms to one side of the equation. In this case, we subtract 3a from both sides.
Step 2: Apply the condition for no real roots
For a quadratic equation Ax2+Bx+C=0 to have no real roots, its discriminant D=B2−4AC must be less than zero. Here, A=2, B=(a−5), and C=(15−3a). We substitute these values into the discriminant inequality.
Step 3: Simplify the inequality
Now, we expand and simplify the inequality obtained from the discriminant. We expand (a−5)2 and distribute the terms in 4(2)(15−3a), then combine like terms to get a simpler quadratic inequality in terms of a.
Step 4: Find the roots of the quadratic inequality
To solve the quadratic inequality a2+14a−95<0, we first find the roots of the corresponding quadratic equation a2+14a−95=0. We can factor this quadratic expression to find the values of a where the expression equals zero.
Step 5: Determine the interval for 'a'
Since the parabola y=a2+14a−95 opens upwards (coefficient of a2 is positive), the expression a2+14a−95 is less than zero between its roots. Thus, the interval for a is (−19,5). This means α=−19 and β=5.
Step 6: Identify integers in the interval and calculate the sum of their squares
The set X consists of all integers x such that −19<x<5. These integers are {−18,−17,…,3,4}. We need to calculate the sum of the squares of these integers. We can use the formula for the sum of the first n squares, ∑k=1nk2=6n(n+1)(2n+1). Since x2=(−x)2, the sum from −18 to 4 can be calculated by summing squares from 1 to 18 and then adding the squares from 1 to 4 and 02. More simply, ∑x=−184x2=∑x=118x2−∑x=14x2+02. No, this is incorrect. The sum is ∑x=−18−1x2+02+∑x=14x2=∑x=118x2+∑x=14x2.
Step 7: Calculate the sum of squares
Using the formula ∑k=1nk2=6n(n+1)(2n+1): \\ For n=18, ∑x=118x2=618(19)(37)=3⋅19⋅37=2109. \\ For n=4, ∑x=14x2=64(5)(9)=2⋅5⋅3=30. \\ The total sum is 2109+30=2139.