High School Math Questions about Unary Quadratic Inequality

a=｛x|-1≤x≤3｝

a∩b=｛x|-1≤x＜2｝

We can use the thinking of function images to let the intersection of y=x +px+q function and x-axis abscissa be less than or equal to -1, and the other is equal to 2 When x=2, 4+2p+q=0 gives the relation.

List the system of equations: 4+2p+q=0 and f(-1) 0 to find the range of values.

The solution interval of the unary quadratic inequality 2x squared - x>0 is expressed as. ,0)u(1/2,+∞

The solution of the unary quadratic inequality x square + x+4 is:

The solution interval of the unary quadratic inequality -x-squared -4x+3 is expressed as.

The solution set of the inequality 2x squared -3x-2>0 is: ( d )-1 2)u(2,+

The solution of the inequality x -2x-3<=0 is a={x|-1 x 3}, because the common part of a and b is -1<=x<2, so when x=2 x +px+q=0, then find q=-2p+4, because x +px+q<0, is a quadratic function, according to the characteristics of the function graph, it is known that it is u-shaped, then this function has two intersection points with the x-axis (y=0), by the function axis of symmetry x=-b 2a, that is, the extreme value of this function y, at this time, the axis of y"'s axis x=1, x +px+q=0 x=-p 2<0,-p 2<1 gets p>-1 2, and then from 4+2p+q=0 and f(-1) 0 to find the range of values p>5 3 q<-2 3, the intersection of the two results gives p>5 3 q<-2 3.

From the functional equation, we can see that when x=0, y=-1

Then only the figure on the top right satisfies this condition.

So only that graph is an image of a function.

And then. x = 1 and x = 1, y = 0. Get.

A + B + A 2 - 1 = 0 (1) A - B + A 2 - 1 = 0 (2), (1) and (2) are added.

Get. 2a + 2a^2 - 2 = 0

That is. a^2 + a - 1 = 0

It can be found with the root finding formula.

a = 1 root number(5)) 2

But the function graph is obvious, the opening is upward, so a>0 then a root has to be discarded.

Take. a = 1 root number(5)) 2

Choosing D is not a question of inequality. It's a quadratic function problem.

This title is a problem of quadratic functions in the ninth grade of junior high school. Pick B.

The analysis is as follows: the axis of symmetry of the parabola in the first and second figures is the y-axis, i.e., x=0By the axis of symmetry equation x=-b2a, which does not correspond to the known condition b>0; The symmetry axis of the parabola in the third and fourth figures is on the right side of the y-axis, indicating the a, b heterotypes.

In the third figure, the opening of the parabola is downward, a < 0, and the known condition b>0 is different; The opening of the parabola in the fourth figure is the same sign as the given condition b>>0, so the image of the quadratic function should be the third graph.

In the third figure, the parabola passes through the origin, there is x=0, y=a2 - 1 =0, then a = 1, but the opening is downward a<0, so there is a = 1 and b

Let the two intersections of the quadratic function y image and the x-axis are (x1,0) and (x2,0) respectively, then x1+x2=-b 2a, and for a b two graphs, b=0 is inconsistent with the condition b>0; Figure c a<0, x1+x2>0, i.e., -b 2a>0 = >b>0;Figure d a>0, x1+x2>0=>b<0, which is inconsistent with the known condition b>0; Therefore, the image corresponding to the function y can only be a c graph, because the function image is (0,0) so a 2=1, and a<0,=>a=-1 The final answer is b

|2x-3|The solution set of >4 is (- 1 2) (7 2,+ the solution set of the inequality x 2+px+q>0 is also this, it is easy to know that -1 2 and 7 2 are unary quadratic equations x 2+px+q=0, and according to the WEIDA theorem, p= - a+b) = - 6, q = a b = - 7 4, p+q= - 31 4

Solution: according to the meaning of the topic; The solution of ax 2+bx+c=0 is x1=2, x2=3, and a<0

x1+x2=-b/a=5, b=-5a

x1x2=c/a=6, c=6a

bx 2+ax+c=-5ax 2+ax+6a>0, i.e., 5x 2-x-6=(5x-6)(x+1)<0 gets the solution set as -1

When a>0 does not conform to a<0, 4a+2b+c=0 9a+3b+c=0 get b=-5a, c=6a, that is, there is a simplification of 5x 2-x-6>0, and you can do the rest by yourself.

x 2-ax+a image respectfully opened to the state Wang, and asked the manuscript to have a unique solution within the range.

Then the equation x 2-ax+a=1 is =0

Get a=2

Discriminant = 4m 2-4(2-m)>0

The product of the two roots is 2-m>0

m<-2 or 1

x∈（-2，-1/2)

The above is the set representation, if you haven't learned, you can't understand it is -2 to draw a diagram, draw a parabola for a one-dimensional quadratic inequality, find two points on the coordinate axis (-2,0) and (,0), draw a parabola with an opening downward, and the x-axis is it.

It depends on the topic'ax^2+bx+c＜0'Equivalent to -(x+2)(x+1 2)<0,', so ax 2-bx+c 0 is equivalent to -(x+2)(x+1 2)>0, i.e., x+2)(x+1 2)<0, so the solution set is -2

1f(x)≥a

f(x)-a

x + ax + 3-a 0 constant established.

Discriminant =a -4(3-a) 0

a²+4a-12≤0

a+6)(a-2)≤0

The axis of symmetry of 6 a 22) f(x)=x +ax+3 is x=-a 2 if -a 2 [-2,2], i.e., the minimum value of f(x) at -4 a 4 is f(-a 2) = 3-a 4 aa +4a-12 0

6≤a≤2-4≤a≤2

If -a 2 2, that is, a -4

Then the minimum value of f(x) is f(2)=4+2a+3 aa -7

7≤a＜-4

If -a 2 2, that is, a -4

Then the minimum value of f(x) is f(-2)=4-2a+3 aa 7 3

aThere is no solution. In summary, when x [-2,2], the range of a, which makes f(x) a constant, is -7 a2

The essence of the two questions is the same, note:

a f(x) is always true, as long as a f(x) is the minimum!

So both problems are actually finding the minimum value of f(x), except that the first problem is to find the minimum value of f(x) on r, and the second problem is to find the minimum value on a given interval. Draw an image to find the minimum value.