Solve the unary quadratic inequality with respect to x

Squaring the unequal sign will remove a x and get b (1-x) 2ab(1-x) + b (1-x).

Then it is simplified to: b(x-1)(x+2a)<=0; Here's the breakdown discussion.

In the first case, b=0, the solution set is r, and in the second case, b>0,a<, the solution set is [1,-2a]; In the third case, b<0,a<, the solution set is (-infinity, 1] and on [-2a, positive infinity); The fourth case; b>0,a>,solution set[2a,1]; Fifth case; b<0,a "solution set (-infinity, -2a] and on [1, positive infinity);

Step-by-step deformation solving:

Right: a x +b (1-x) a x +b (1-x) +2abx (1-x).

Elimination A x, Approximate (1-x): B b (1-x)+2abx, x≠1

Approximate b: b b-bx+2ax, b≠0

Shift item: bx-2ax 0

Discussion on a case-by-case basis:

If b-2a 0, i.e., a b and b≠0, then x 0 (the original formula also holds when x=1);

If b-2a=0, i.e., a= b and b≠0, then x is any real number;

If b-2a 0, i.e., a b and b≠0, then x 0;

If b=0, then x is any real number.

That's it.

a²x²+b²(1-x)+≥ax+b(1-x)]²a²x²+b²(1-x)≥a²x²+2axb(1-x)+b²(1-x)²

b (1-x)[1-(1-x)] 2axb(1-x) seems to be discussed on a case-by-case basis.

The unary quadratic inequality is found as follows:

Original question] Solve the following inequality:

2x2＋3x＋2<0。

Solution] (1) 2x2 3x 2<0, that is, 2x2 3x 2>0.

The two real roots of equation 2x2 3x 2 0 are x1 , x2 2.

So the solution set of 2x2 3x 2>0 is , that is, the solution set of the original inequality is .

The concept of unary quadratic inequalities:

An inequality that contains only one unknown and the highest number of unknowns is 2 is called a unary quadratic inequality.

The set of solutions for unary quadratic inequalities:

The value of x that makes a certain unary quadratic inequality true is called the solution of the unary quadratic inequality, and the set of all the solutions of the unary quadratic inequality is called the solution set of the unary quadratic inequality.

General steps and precautions for solving unary quadratic inequalities:

1. Chemistry. The inequality is deformed so that one end is 0 and the quadratic coefficient is greater than 0, i.e., ax2 bx c 0 (a 0), ax2 bx c 0 (a 0).

2. Judgment. Calculate the discriminant of the corresponding equation.

3. Seek. When δ 0, find the root of the corresponding one-dimensional quadratic equation, or indicate whether the equation has a real root according to the discriminant formula.

4. Write. According to the image of the corresponding quadratic function, the solution set of inequalities is written by using "greater than taking both sides, less than taking middle".

Problems that should be paid attention to when solving unary quadratic inequalities:

1. When solving the quadratic inequality, the coefficient of the quadratic term should be turned into a positive number first.

2. When the quadratic coefficient contains parameters, the sign of the parameter will affect the solution set of the inequality, and do not forget that the quadratic coefficient is zero.

3. To solve the problem of the constant establishment of a one-dimensional quadratic inequality, we should pay attention to the sign of the coefficient of the quadratic term.

4. The endpoint of the solution set of the unary quadratic inequality is the same as the root of the corresponding unary quadratic equation and the corresponding quadratic function image and the abscissa of the intersection point of the x-axis.

Summary. The general solution to the quadratic inequality is to make it equal to 0 and then combine it according to the image number.

Solve a unary quadratic inequality.

The plus sign in the lower right corner can be patted**.

Send us the original question.

The general solution to the quadratic inequality is to make it equal to 0 and then combine it according to the image number.

Start with the original question. If you can't turn it on**, take a photo with your mobile phone, and then the plus sign in the lower right corner can be sent**.

6xx－16x＋10＜0

OK If you want to draw an image.

Because drawing an image is the fastest.

First of all, it is based on the three-point method. Combine the properties of quadratic equations.

I'll just say when he's equal to 0 and draw a simple diagram based on the direction of the opening.

Let's see when he is equal to 0, and then draw a simple diagram according to the direction of the opening.

Then, according to the image, you can intuitively judge that it is at 1 5 3.

It is Xiaoyu who leads, so it is said that x belongs to 1 5 3.

Be sure to pay attention to the open range.

If the equation can be simplified, it should be simplified first.

Quadratic functions have roots, because you're only halfway there.

This function has two with one 1 and one 2.

Plus it is open upwards.

So he's definitely between one and two.

So it is greater than or equal to 1<=2.

Do this kind of function-wide question.

It is advisable to draw an image.

The combination of numbers and shapes is the most intuitive and convenient.

(x-3a)(x+a)>0

x1=3a x2=-a

If a>0

Then x<-a or x>3a

If a>0

Then x<-a or x>3a

If a=0, then x<>0

(Represents not equal).

To solve the inequality, first find the two roots, and then look at the quadratic coefficient of the quadratic function, when the quadratic coefficient is greater than 0.

If it is followed by a greater than sign.

Then the solution of the inequality is.

xb (a if it is followed by a less than sign.)

Then the solution of the inequality is.

a, so the two solutions we find are positive and negative in the case of an unknown size!

(x-3a)(x+a)>0 1, if a<0 then x<3a or x>a

2. If a=0 then x is any real number.

3. If a>0 then x<-a or x>3a

x 2-2ax-3a 2=(x-3a)(x+a)>0 when a<0, x<3a or x>a

When a=0, x is not equal to 0

When a>0, x<-a or x>3a

x2-2ax-3a2＞0

x-3a)(x+a)>0

X-3a is timing with x+a, and x-3a >0

x+a>0

At this time, when a>0, the solution is x>3a

When a=0, the solution is x>0

When a<0, the solution is x>-a

When x-3a and x+a are both negative, x-3a < 0

x+a<0

In this case, when a>0, the solution is x<-a

When a=0, the solution is x>0

When a<0, the solution is x<3a

In summary, when a>0 x>3a or x<-a

When a=0, x=0

When a<0, x<3a or x>-a

From the root finding formula, we can find x=a or x=1

When a<1, the solution is a1, and the solution is 1

x（x-1）≥0

x≥0，x-1≥0

or x 0, x - 1 0

Solve the group of inequalities, and solve x 1 or x 0

The set of solutions to the original inequality is x 1 or x 0

Quadratic functions!! If the opening of the parabolic trace is facing upwards, it is greater than (equal to) the large root or smaller than (waiting for a good file and stupid in) the small root!

1.Original = (x-1)(x-1) 0

Solve x 1 or x 1

2.Original = (x-5)(x+2) 0

Solution: -2 x 5

5x²-2x+8=5(x-1/5)²-5*(1/5)²+8=5(x-1/5)²+39/5

Regardless of the value of x, the above equation is greater than 0, so the solution set is r

The following question can also be solved in the same way.

If you still have questions, ask me!

(1) y=2x +ax+2 opening up, the minimum value is -(a -16) 8, -44 or a<-4, x>(-a+sqrt(a -16)) 4 or x<(-a-sqrt(a -16)) 4, sqrt is the following sign.

2) The same opening is upward, the minimum value is -((a +a) -4a ) 4=-(a -a) 4, and the total intersection point is with the x-axis.

Intersection with x-axis ((a +a)-(a -a)) 2=a, (a +a)+(a -a)) 2=a

If a>a i.e. 1a or x1 or a<0 then xa

It's a two-dimensional right... Treat a as a constant, calculate the range of x values separately, and then combine the results of the two equations. Solution.

First of all, oh, it's a one-dimensional quadratic inequality, not a binary one-time, right?

1）x²+（a/2）x+1>0

x²+a/2x+a²/16-a²/16)+1>0(x+a/4)²>a²/16-1

When a 16-1 0.

1,) x+a 4<-under the root number (A 16-1) x <-under the root number (A 16-1)-A 4

2, x+a 4> (A 16-1) x > (A 16-1)-A 4

When a 16-1 0.

x is any real number.

2）(x-a)(x-a²)>0

x-a>0;x-a >0 to get x>a and x>a or x-a<0; x-a <0 to x