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Inequality of quadratic function (19 20:56:43) Given the quadratic function f (x) = AX2 + BX + 1 (a, B ∈ R, a > 0), let the two real roots of the equation f (x) = x be X1 and x2 (1) If x1
Given the quadratic function f (x) = ax ^ 2 + BX + 1 (a > 0), let the two real roots of the equation f (x) = x be X1 and X2 (1) if a = 2 and X1 1 + 2 ^ (3 / 2) or B < 1 - 2 ^ (3 / 2)
The inequality of solution about X: AX-1 / x + 2 > 0
The inequality (AX-1) / (x + 2) > 0 with respect to X
② When a ≠ 0, X1 = 1 / A, X2 = - 2
1) When a ＞ 0, x1 ＞ X2, the solution of inequality is: X ＞ 1 / A or X ＜ - 2
2) When a ＜ - 1 / 2, x1 ＜ X2, the solution of inequality is: - 1 / a ＜ x ＜ - 2
3) When - 1 / 2 ＜ a ＜ 0, x1 ＞ X2, the solution of the inequality is: - 2 ＜ x ＜ - 1 / A
That's the answer online, but I don't know why
There is less discussion about the case of a = 0, when a = 0, x0 and a
Solving inequality (X-2) (AX-2) > 0
If a = 0, then inequality (X-2) (AX-2) > 0 is equivalent to - 2 (X-2) > 0, the solution is: x2 or x0 is equivalent to (X-2) &# 178; > 0, the solution is: X ≠ 2;
When 02 / A or X
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