If a satisfies the equation x2 + (1-2a) x + A2 = 0 at the same time, the root of the equation is judged______ ．

If a satisfies the equation x2 + (1-2a) x + A2 = 0 at the same time, the root of the equation is judged______ ．

When the image of ∵ positive scale function y = (a + 1) x passes through the second and fourth quadrants, the ∵ a + 1 ＜ 0, i.e., a ＜ - 1, ∵ △ = (1-2a) 2-4a2 = 1-4a, ∵ a ＜ - 1, ∵ 1-4a ＞ 0, i.e., △ 0, the equation has two unequal real roots

How to judge whether the inverse scale function image is in the first three quadrants or the second four quadrants

According to the positive and negative of K
One, three
Negative, two or four

If the image of the function y = - ax + 2a-1 does not pass through the first quadrant, the value range of a is obtained

First understand that this is a linear function
Because it doesn't go through 1 quadrant
It can be through 2,3,4 quadrant or 2,4 quadrant
So - A is less than 0
That is, a is greater than 0
And the constant 2a-1 should be less than or equal to 0
So 0