What if the discriminant of a quadratic function is a quadratic function? For example: F (x) = x & # 178; + (2m-1) x + 2m & # 178; + 1 △=（2m-1）² -4（2m²+1）

What if the discriminant of a quadratic function is a quadratic function? For example: F (x) = x & # 178; + (2m-1) x + 2m & # 178; + 1 △=（2m-1）² -4（2m²+1）

Discriminant is to see its relationship with 0
So we need to simplify △ first
（2m-1）² -4（2m²+1）
=4m²-4m+1-8m²-4
=-4（m²+m+1/4）-2
=-4(m+1/2)²-2
Let's begin with the discussion

If the discriminant is less than zero, then the value of quadratic function is greater than 0. Is this conclusion correct? Why?
The original problem is as follows: find the definition field and value field of F (x) = √ (5x ^ 2 + 8x + 5).
∵ the discriminant is less than zero. That 5x ^ 2 + 8x + 5 is greater than zero is tenable
And 5x ^ 2 + 8x + 5 ≥ 9 / 5? )
The definition field of F (x) is r, and the value field is (3 / 5) √ 5 ≤ y

In quadratic function of one variable: F (x) = ax ^ 2 + BX + C
If a > 0, then f (x) is a parabola with the opening upward. If there is no intersection between the function and X axis, then the function is always greater than 0
If a = 0
To solve this inequality, we first solve the equation 5x ^ 2 + 8x + 5 = 0, the equation has no solution, and because the known function of 5 > 0 is a parabola with the opening upward, it can be judged that the inequality (5x ^ 2 + 8x + 5) > = 0 holds no matter x is any value
So the domain of F (x) = √ (5x ^ 2 + 8x + 5) is (- ∞, + ∞), which is the real number set R
Let's find the range of values. We know that the maximum value of (5x ^ 2 + 8x + 5) is + ∞. Now we find its minimum value
When x = - B / 2A = - 8 / (5 * 2), there is a minimum, and the minimum of (5x ^ 2 + 8x + 5) is 9 / 5
So the range of F (x) is [√ (9 / 5), + ∞)