High school math solves quadratic inequality problems, the more detailed the better.

The solution set of the quadratic inequality ax +bx+c<0 (>0) is based on.

The intersection of the quadratic function image and the x-axis is observed.

The solution set of inequalities is closely related to quadratic equations and quadratic functions.

The solution set for x + ax + b 0 is [-1,3].

Then x1 = 1 and x2 = 3 are the equations.

x +ax + b = 0.

According to Veda's theorem:

x1+x2=-a=-1+3=2

x1x2=b=-3

a=-2,b=-3

To solve it as an equation, -1 and 3 are its two roots, using Veda's theorem -1*3=b=-3 -1+3=2=-a

So a=-2 b=-3It's the easiest way to do it!

If you still don't understand, you can bring a=-2, b=-3 back to the inequality, solve it yourself, see if the value of x is between -1 and 3, and slowly understand, of course, you can also draw a quadratic image to understand.

Solution: As can be seen from the problem, the two roots of the equation are -1 and 3, which is known by Vedd's theorem: -1+3=-a, -1*3=b

Solution: a=-2, b=-3

It can be analyzed that (x+1)(x-3)=x=x, squared +ax+b, that is, a=-2, b=-3

By the title b=[-1,3].

So -1 and 3 are the two real roots of the equation x2+ax+b=0, which is obtained by Vida's theorem a=-2 and b=-3

b=[-1,3]

The first step is to see if the sign in front of the quadratic term is positive, if it is negative, then move the term to the other side of the unequal sign, become positive, and the other side is 0, the second step, judge the situation of the corresponding equation solution, you can make the unequal sign become an equal sign, there are two main situations for judging the solution, the first choice is factorization, there must be a solution for those who can be factored, and those that cannot be factored are judged by Delta, the third step is to draw an image of the corresponding quadratic function according to the situation of the solution, because the sign in front of the quadratic term is positive, Therefore, generally draw a parabola with an opening upward, the fourth step, according to the unequal relationship to observe the image to judge the solution set, such as 0, then look at the value range of x reflected by the image above the x-axis, 0 then look at the image below, generally if the equation has two solutions, then remember the formula: greater than take both sides, less than take the middle, you can directly write the inequality solution set. For example, if the solution of the equation is 1 and 3, and the inequality is eventually 0, then the solution of the inequality is x 3 or x 1, if the inequality is 0, then the solution is 1 x 3, and so on.

Solution: m = - 3 2

n = 3/2

Analysis: (1) The solution to an inequality of the form ax + bx + c 0 (a 0) is: x x1 or x x2

x1 and x2 are the two roots of the equation ax + bx + c = 0, and x1 x2).

2) The solution to an inequality of the form ax + bx + c 0 (a 0) is: x2 x x1

x1 and x2 are the two roots of the equation ax + bx + c = 0, and x1 x2).

In response to your question, the solution set of inequalities with the right end greater than zero is in the form of "x2 x x1", which requires multiplying both sides of the original inequality by (1) to turn it into an inequality with the right end less than zero.

The solution set of the inequality mx2+nx+3 greater than 0 is (-1,2), indicating that the solution set of the inequality (-m)x -nx-3 0 is (-1,2).

It shows that the two roots of equation (-m) x -nx - 3 = 0 are -1 and 2.

From the relationship between the root and the coefficient, it is known:

The sum of the two: (-1) +2 = -n m

n = m ——

The product of two roots: (-1) 2 = 3 m

2m = 3 ——

Obtained by solution: m = - 3 2 ; n = 3/2

Good luck with your studies!

The endpoints of the solution set of quadratic inequalities are the two roots corresponding to the quadratic equation, so the simplest method is Vedica's theorem: the product of the two roots c a=3 m=-2, so we get: m=-3 2;

The sum of the two roots, -b a=-n m=1, so we get: n=-m=3 2;

i.e. m=-3 2, n=3 2;

I hope it can help you, if you don't understand, please hi me, I wish you progress in your studies!

There are three scenarios.

1. There is an intersection between the curve and the x-axis, then b 2-4ac=0, (where a is not the a, in order not to be confused, I use n instead of n), n=+-4, when x=-b 2a=-n 2, then the formula is 0, which is also the minimum value, at this time x belongs to r, and is not equal to.

n 2, x 2 + nx + 4>0 is established.

2. When the curve has no intersection with the x-axis, b 2-4ac<0, that is, -40, that is, n>4 or n<-4, because the curve opening is upward, you solve.

x 2 + nx + 4 = 0 (using the root finding formula, x1 and x2 are solved), at this time x must be larger than the large solution and smaller than the small solution before the equation holds.

You didn't understand the question under what circumstances x should meet x 2+ax+4>0, (a r).

That's what it means, isn't it coming out of the formula for finding the root? I'll write you a detail later.

First of all, it is necessary to discuss whether the equation has a real number solution, that is, to discuss a 2-4*4 in three cases: less than zero, equal to zero, greater than zero can get the answer.

Solution: The solution set of (m 2) x 2 (m 2) x 4 0 is an empty set, that is, the solution set of (m 2) x 2 (m 2) x 4>0 is the whole real number r, so it is divided into two cases.

1) When m-2=0, that is, m=2, 4>0 meets the requirements (2) When m-2≠0, then according to the requirements of the question, m-2>0 (opening up) 0 (no intersection with the x-axis) i.e.

m-2>0

[2(m2)] 4(m-2)*4<0 is solved to get 2 in summary, then the value range of m is [2,6].

Having said that it is a quadratic inequality, m cannot be equal to 2. For a unary quadratic equation less than zero, then the opening is upward and has no intersection with the x-axis, so b 2-4ac < 0

At 22 o'clock, the square of b is used to subtract the binary inequality of 4ac train with respect to m to solve 2

Let t(x)=ax +4x+a-3>0

To make the f(x) range r.

When a=0, 4x-3>0 is solved to x>3 4

When a≠0, a>0; =16-4a(a-3)>=0 to get the value of 0, so a is [0,4].

The value range is r, and ax2+4x+a-3 can be defined in "0".

a=0,4x-3>0 OK.

a = 0, should have δ< = 0

The ground floor is right.

value range r, make sure that ax 2+4x+a-3>0 also make sure that ax 2+4x+a-3 takes the value of (0, +oo).

When a=0, ax 2+4x+a-3=4x-3 satisfies the requirements.

When A < 0, round off A

A>0, ax 2+4x+a-3 = a(x+2 a) 2+(a-4)(a+1) a

Must meet (a-4)(a+1) a>0, =>(a-4)(a+1)>0

a<-1 , a>4 , this is the correct solution.

If the discriminant formula is greater than the range of the value of zero, the true number is greater than zero, but it is not guaranteed to be (0,+oo).

f(x)=ax2+bx+c, satisfying f(0)=f(1)=0,, its minimum value is -1 4, and the solution gives a=1, b=-1, c=0, so the function is f(x)=x 2-x

From the meaning of the title, we know that f(x)+f(1 x) (x+1 x)*lnm x is not equal to 0 (1) constantly, and when x>0, (1) can be transformed into lnm<=[f(x)+f(1 x)] (x+1 x).

Let the function g(x)=[f(x)+f(1 x)] (x+1 x)=[x 2-x+x (-2)-1 x] (x+1 x).

The function g(x) is derived and its monotonic interval is obtained, g'(x)=(x+1)(x-1)(x 4+4x 2+1) [x 2*(x 2+1)].

It is obtained that when x>0 the function obtains the minimum value at x=1 and g(1)=0

Since lnm<=g(x) and the minimum value of g(x) is 0, lnm<=0 and m<=1.

When x<0, (1) can be transformed into lnm>=[f(x)+f(1 x)] (x+1 x).

The derivative of the function g(x) is the same as above, and when x<0 the function g(x) obtains the maximum value at x=-1, and g(-1)=-2

Since lnm>=g(x) and the maximum value of g(x) is -2, lnm>=-2 and m>=e (-2).

In summary, when e(-2)<=m<=1, there is always f(x)+f(1 x) (x+1 x)lnm for any function.

Let x+1 x=t, known inequality can be reduced to t 2-t(1-4lnm)-2>=0, when x>0, t>2, only the value of m when t 2-t(1-4lnm)-2=0, t=2 can be obtained. m=1, i.e., m>1, the inequality is known to hold constantly.

When x<0 and t<-2, only t 2-t(1-4lnm)-2=0 and t=-2 are required. lnm=3 8, which is m< e (3 8), the inequality is known to hold constantly.

x+2≥0x^2-4)^2≤(x+2)^2x+2)^2(x-2)^2≤(x+2)^2x+2)^2(x^2-4x+3)≤0

1) When x+2=0, lead hail x=-2, which is eligible;

2) When x+2≠0, Huaifan x 2-4x+3 01 x 3, so the solution of the original inequality is 1 x Yu Hao 3 or x=-2