If the inequality ax ^ 2 + BX + C0

If the inequality ax ^ 2 + BX + C0

The solution of ax ^ 2 + BX + C = 0 (a is not equal to 0) is x = 2,3; the coefficient obtained by Weida's theorem is a = 1, B = - 5, C = 6, which is brought into the following inequality, and the solution is X5 / 6
If the solution set of quadratic inequality ax ^ 2 + BX + 2 > 0 is (- 1 / 2,1 / 3), then the value of a + B is?
The solution set of quadratic inequality ax ^ 2 + BX + 2 > 0 is (- 1 / 2,1 / 3),
So the root of the corresponding equation is X1 = - 1 / 2, X2 = 1 / 3
Using the relationship between root and coefficient, we get X1 + x2 = - B / a = - 1 / 2 + 1 / 3 = - 1 / 6
x1*x2=2/a=-1/2*1/3=-1/6
So a = - 12, B = - 2. So a + B = - 14
I don't know if you understand
Quadratic inequality of ^ 2 / x + 11
From the meaning of the title
(x+1/2)(x-1/3)>0
The results are as follows
x^2+(1/2-1/3)x-1/6>0
x^2+1/6x-1/6>0
6x^2+x-1>0
The two sides are multiplied by - 1, and the unequal sign is reversed
-6x^2-x+1
First, we look at the discriminant B ^ 2-4a > 0 to get B ^ 2 > 4a
Then write out the two roots
Let one equal 1 / 3 and the other equal - 1 / 2
Let's use the two alternative formulas to judge the two cases.
Finally, it is calculated that:
a= -6
b= 1
Have you studied the relationship between root and coefficient?
x1+x2=b/a
x1*x2=1/a
Use 1 / 3, - 1 / 2 generation of x1.x2
Quadratic function f (x) = ax ^ 2 + BX + C (a > 0) f (1) = - A / 2 prove that there is at least one zero point between the interval (0,2)
f(1)=a+b+c=-a/2;
f(0)=c;
f(2)=4a+2b+c=2a+2(a+b+c)-c=a-c;
a> So if f (1) 0, f (0) = C > 0, then f (0) * f (1)
In fact, it's very simple to do this kind of problem. First, we should grasp the f (1) = - A / 2 conditional connection (0, 2) and discuss it by combining with the characteristics of the concept of zero
f(1)=a+b+c=-a/2; f(0)=c;
f(2)=4a+2b+c=2a+2(a+b+c)-c=a-c;
a> So if f (1) 0, f (0) = C > 0, then f (0) * f (1) 0; then f (0) * f (1)
Let f (2x-1) = 3x-1 / x + 2, then f (5) =?
2x-1=5
2x=6
X=3
f(5)=3x3-1/3+2 =8/5
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Let 2x-1 = 5
Then x = 3
therefore
f(5)= 3*3 - 1/3 + 2
=9+2- 1/3 = 32/3
f(5) = (3*3 - 1) / (3 + 2) = 8/5
The two answers are because I don't know your way of breaking sentences.
It is known that the quadratic function f (x) = ax ^ 2 + BX + C satisfies √ 2A + C / √ 2 > b, and C
My answer is d
From √ 2A + C / √ 2 > b, it is reduced to: 2a-b + C > 0, that is, when x = √ 2, f (x) is greater than 0, so the intersection of parabola and X √ 2 axis is from 0 to √ 2, so it is more on: (0,2)
The answer is B: don't guess, OK??? depressed
Given the function f (2x + 1) = 3x + 2, find f (x)
How to determine who should be set as t in such a problem?
Let 2x + 1 = a, x = (A / 2) - (1 / 2)
f(a)=(3a/2)-(3/2)+2=(3a/2)+(1/2)
Let x = 2x + 1, then f (x) = 3 (2x + 1) + 2
=6X+5
Given that one zero point of quadratic function f (x) = x ^ 2 + BX + 4 (B ∈ z) is on (0,2) and the other zero point is on (3,5), then B=
If f (0) > 0, then f (2) is obtained from zero point on (0,2)
Axis of symmetry - B / 2 = 2.5
B = - 5
b=-5
By using the distribution of roots, the satisfaction can be known from the image
f(3)0
b∈Z
B = - 5
Given f (2x + 1) = 3x-1, find the analytic expression of function f (x)
Let 2x + 1 = M
Then x = (m-1) / 2
So f (2x + 1) = f (m) = 3x-1 = 3 * [(m-1) / 2] - 1 = 3m / 2-5 / 2
That is, f (x) = 3x / 2-5 / 2
=3（2x+1）-1
=6x+2
F (x) = 3 / 2x-5 / 2
A zero point of quadratic function f (x) = ax ^ 2 + BX + C is - 1, [f (x) - x] * [f (x) - (x ^ 2 + 1) / 2]
(-1,0),a-b+c=0
[f(x)-x]*[f(x)-(x^2+1)/2] =(ax^2+bx+c-x)*(ax^2+bx+c-x^2/2-1/2)