If the proposition "existence x belongs to R, 2x-3ax + 9 ＜ 0" is a false proposition, how to find the value range of real number a? The answer is - 2 root 2 to 2 root 2

If the proposition "existence x belongs to R, 2x-3ax + 9 ＜ 0" is a false proposition, how to find the value range of real number a? The answer is - 2 root 2 to 2 root 2

The proposition is equivalent to any x belonging to R, 2x2-3ax + 9 > = 0, that is, b2-4ac

Proposition "being X. &; R, 2x. - 3ax. + 9"

Its negative proposition "&#; X ∈ R, 2x2-3ax + 9 ≥ 0" is a true proposition, which is also a common "constant hold" problem, only △ ≤ 0
The negative proposition of the original proposition is "&; X ∈ R, 2x2-3ax + 9 ≥ 0", and it is true,
If the value of quadratic function with opening upward is greater than or equal to 0, it will be true,
Only △ = 9a2-4 × 2 × 9 ≤ 0, the solution is: - 2 ≤ a ≤ 2
So the answer is: [- 2,2]

If the proposition "彐 x ∈ R, 2x ^ 2-3ax + 9"

From 2x ^ 2-3ax + 9 ＜ 0, 2 〔 x ^ 2 - (3a / 2) x + (3a / 4) ^ 2 〕 - 2 × (3a / 4) ^ 2 + 9 ＜ 0,
∴2（x－3a/4）^2－9a^2/8＋9＜0.
Obviously, when - 9A ^ 2 / 8 + 9 ≥ 0, 2x ^ 2 - 3ax + 9 < 0 is a false proposition
From - 9A ^ 2 / 8 + 9 ≥ 0, we can get a ^ 2 ≤ 8, A-2 √ 2 ≤ a ≤ 2 √ 2
The value range of a satisfying the condition is [- 2 √ 2,2 √ 2]