Given that the solution of the inequality ax & # 178; + BX + C < 0 (a ≠ 0) is x < 2, or x > 3, find the solution of the inequality BX & # 178; + ax + C > 0 Especially, why is the root of inequality x = 2,3? Shouldn't it be a fixed value?

Given that the solution of the inequality ax & # 178; + BX + C < 0 (a ≠ 0) is x < 2, or x > 3, find the solution of the inequality BX & # 178; + ax + C > 0 Especially, why is the root of inequality x = 2,3? Shouldn't it be a fixed value?

The graph of y = ax & # 178; + BX + C is a parabola with an opening up or down (depending on a and b). The solution satisfying ax & # 178; + BX + C < 0 includes all x values under the curve, so it is not a fixed value
X ＜ 2 and X ＞ 3 are the solutions of the inequality, which indicates that the curve is open up, and through (2,0) and (3,0), the lowest point of the curve is x = 2.5
Given that the solution of inequality ax & # 178; + BX + C ＜ 0 (a ≠ 0) is x ＜ 2 or X ＞ 3, find the solution of inequality BX & # 178; + ax + C ＞ 0
I found the answer on the Internet.
X0
(x-2)(x-3)0
-5x + 6 > 0 and X + 1 > 0 get - 1
The solution of known inequality ax & # 178; + BX + C0?
The solution set of AX2 + BX + C ＜ 0 is {x | x ＜ 2 or X ＞ 3},
The roots of ax & # 178; + BX + C = 0 are 3 and 2, and a < 0
According to Weida's theorem: 3 + 2 = - B / A
3×2=c/a
The solution is b = - 5A, C = 6A
Then the inequality BX & # 178; + ax + C > 0 can be reduced to:
-5ax ²+ax+6a>0
That is 5x & # 178; - x-6-1
Given that the solution set of inequality ax & # 178; + BX + 2 > 0 is {x | - 1 / 2 ＜ x ＜ 1 / 3}, find the solution set of inequality ax & # 178; - BX + 2 > 0
-1/2*(1/3)=2/a
a=-12
-1/2+1/3=-1/6=-b/a
1/6=b/(-12)
b=-2
-12x²+2x＋2＞0
6x²-x-1
Ax & # 178; + BX + 2 = 0 are - 1 / 2,1 / 3
-1/2+1/3=-b/a，-1/6=2/a
a=-12，b-2
The solution set of inequality - 12x & # 178; + 2x + 2 > 0 is {x | - 1 / 3 ＜ x ＜ 1 / 2},
Given that the solution set of inequality AX2 + BX + 2 ＞ 0 is {x | - 12 ＜ x ＜ 13}, then the solution set of inequality 2x2 + BX + a ＜ 0 is {x | - 12 ＜ x ＜ 13}______ .
The solution set of the inequality AX2 + BX + 2 ＞ 0 is {x | - 12 ＜ x ＜ 13}, and the inequality 2x2 + BX + a ＜ 0 is transformed into 2x2-2x-12 ＜ 0, that is, x2-x-6 ＜ 0, and the solution set of the inequality 2x2 + BX + a ＜ 0 is (- 2, 3). So the answer is: (- 2, 3)
If the solution of the inequality ax & # 178; + BX + C > 0 is 3 < X
The solution of ∵ ax & # 178; + BX + C > 0 is 3 < X
Because the solution is 3 < x0
So - (x-3) (X-5) > 0
So - x ^ 2 + 8x-15 > 0
So a = - 1, B = 8, C = - 15
cx²+bx+a
Given that the solution set of inequality ax & # 178; + BX + C > 0 is (- 2,3), then inequality CX & # 178; + ax-b
Because the solution set of ax & # 178; + BX + C > 0 is a closed interval (- 2,3), so a < 0, and the solution of the equation AX & # 178; + BX + C = 0 is X1 = - 2, X2 = 3. According to Weida's theorem X1 + X2 = - B / a = 1, so B = - A, x1x2 = C / a = - 6, there is C = - 6A. So CX & # 178; + ax-b
Given that the solution set of inequality ax & # 178; + BX + C > 0 is {2 < x < 3}, find the solution set of inequality CX & # 178; - BX + a > 0
{X>1/2}
If the solution set of quadratic inequality ax ^ 2 + BX + C > 0 is (α, β) (α > 0), then the solution set of inequality CX ^ 2 + BX + a > 0 is (α, β)
Analysis: because the solution set of the quadratic inequality ax ^ 2 + BX + C > 0 is (α, β), we know that A0, then β > α > 0, so - B / a > 0, C / a > 0 then B0. At this time, we can set the two roots of the quadratic equation CX ^ 2 + BX + a = 0 corresponding to the inequality CX ^ 2 + BX + a > 0 as m, n because C > 0, so the relationship between the two and the coefficient is: M + n = - B / C = (- B / a
The solution set of ax ^ 2 + BX + C > 0 is (α, β) (α > 0),
Then a (x - α) (x - β) > 0 is equivalent to AX ^ 2 + BX + C > 0
And A0
c=a((αβ)0
A α β x ^ 2-A ((α + β) x + a > 0, so α β x ^ 2 - (α + β) x + 10) is obtained,
Then a (x - α) (x - β) > 0 is equivalent to AX ^ 2 + BX + C > 0
And A0
c=a((αβ)0
A α β x ^ 2-A ((α + β) x + a > 0) is obtained, so α β x ^ 2 - (α + β) x + 1
If the solution set of quadratic inequality ax ^ 2 + BX + 3 > 0 is (- 3,1 / 2), then the value of a + B is ()
If the solution set of quadratic inequality ax ^ 2 + BX + 3 > 0 is (- 3,1 / 2), then the value of a + B is ()
If the solution set of quadratic inequality ax ^ 2 + BX + 3 > 0 is (- 3,1 / 2), then the value of a + B is (- 7)
∴-3+1/2=-b/a;
-5/2=-b/a;
-3×1/2=3/a;
∴a=-2;
b=-5;
∴a+b=-2-5=-7;
I'm very glad to answer your questions. Skyhunter 002 will answer your questions
If you don't understand this question, you can ask,
The solution set of inequality ax & # 178; + BX + 3 > 0 is (- 3,1 / 2)
It is shown that the two solutions of the equation AX ^ 2 + BX + 3 = 0 are - 3 and 1 / 2
-3+1/2＝-b/a
-3*1/2=3/a
a=-2
B=5
So: a + B = 3