High School Root Distribution Problem Master Advanced, High School Mathematics Root Distribution Pro

I can't write so much, but this is a method taught by the top class teacher of Chongqing Bashu Middle School that will never be missed.

The main points are: 1. The axis of symmetry is in the left, middle and right of the interval. (Consider whether a is 0) 2. List the magnitude relationship between f(a), f(b) and 0 in combination with the graph, and consider solving the inequality groups separately when they are in the interval, and synthesize them. Upstairs there is a possibility of leakage.

To prevent confusion between the a,b interval and the coefficient, [p,q] is used to refer to the interval below:

1.delta>=0, and.

f(p)f(q)<=0 (this is one condition) or af(p)>=0, af(q)>=0, p=<-b (2a)<=q (this is two conditions).

2.delta>=0, and.

af(p)>0, af(q)>0, p<-b (2a)=0, and af(p)<0, af(q)<0

The first is to discuss whether a is 0, if it is 0, b must be greater than 0, that is, when bx+c=0 has a root. x=-c/b.If a is not 0Use the discriminant expression of the root to solve it, b 2-4ac> or = 0.

The second one: the same as before, that is, when a is not 0, the discriminant formula b 2-4ac > 0, and the solution is obtained.

The third: the same as before, that is, when a is not 0, the discriminant b 2-4ac < 0, and the solution is obtained.

1.The discriminant formula is greater than or equal to 0, the axis of symmetry is between a b, and the sum of the two is less than or equal to half of b minus a

2.If the discriminant formula is greater than 0, the axis of symmetry is between a b, and the sum of the two is less than or equal to half b minus a3The discriminant formula is less than zero. Well.

The first question is f(a).f(b)<0

The second question is B 2-4AC>0 A < B 2A0

The third question is -b 2a0 or -b 2a>b, f(a)f(b)>0

Typical categorical discussions:1When x>=0 does not match the title, it is discarded.

0 then the equation is deformed to -x-ax-1=0 x=-1 (1+a), and because x<0 is solved a>-1

The equation will not have 2 negative roots.

So the meaning of the title is 1) there may be only one root, and it is negative, and 2) there may be 2 roots, one of which is negative.

So got. x<0:-x-ax-1=0

x=-2/(1+a)<0

then a>-1

Question 1: Since one of them is in the interval (-1,0) and the other is in the interval (1,2), so his axis of symmetry should be in (0,1), you can draw a diagram to think about it, so you can directly solve that m is also in the scum pose of (-1,0).

Question 2: Since both roots of the equation are in the interval (0,1), so the symmetry axis of Sou Liang is (0,1) and f(0)>0,f(1)>0, and the discriminant formula is greater than 0, and from these four conditions, it can be concluded that m is still between (-1 2,1-root number 2).

If there are two positive roots, then 0 must be guaranteed, and according to Veda's theorem, x1+x2=-b a 0, x1*x2=c a 0(It is guaranteed that there are two roots of the equation, and Vedic theorem is to ensure that the two roots are positive numbers, because the product of two numbers is greater than zero, then the two numbers must have the same sign, either positive or negative, but at the same time the sum of the two roots is greater than zero, so the two must be positive).

Or directly find the two, find the root formula x=(-b b -4ac) 2a, and directly substitute it into the calculation.

The idea of combining numbers and shapes:

x 2 + 1 = -mx represents the intersection of the parabola y=x +1 and the straight line y=-mx over the origin.

The parabola is known to pass through the points (1,2).

Then, it is necessary to satisfy that they have an intersection in (0,1).

Then the slope -m of the straight line y=-mx is greater than the slope of the line connecting the point (1,2) with the origin (0,0), i.e., -m 2

So, m -2

ps: The derivative in it is to find the slope of the tangent at any point on the parabola - it is estimated that you have not learned the "derivative" yet, so you can't understand it for the time being!!

The landlord's so-called combination of numbers and shapes is a method of combining numerical values with graphics and images to analyze the problem.

The derivative in PS uses the knowledge that currently belongs to university mathematics. At present, there is no introduction to this knowledge in high school textbooks, so you don't need to worry too much. Derivation is only a means of verifying the answer to the problem, and cannot be used in the solution of the high school exam.

x is between (0,1).

Make y=x +1 intersect with y=-mx.

As you can see, if you take the same x, the value of y=-mx is greater than y=x +1.

i.e. 2<-m·1

So, m -2

x 2+1>-mx when x = 0

When x=1 is relative, x 2+1<-mx gives m<-2, otherwise there would be no real root.

ps means to think of it as two functions with the same defined domain.

The two x1 and x2 of the equation x 2 + (m-3) x + m = 0 are both negative, ==> =(m-3) 2-4m=m 2-10m+9>=0, x1+x2=3-m<0, x1x2=m>0, and the value range of m>=9 or m<=1, m>3, m is [9,+

Variant 1 Both roots are positive roots, the above x1+x2=3-m becomes 》0, the remainder remains unchanged, and the value range of m is (0,1).

Variant 2 Both are within (0,3) and on the basis of Variant 1, increase.

x1-3+x2-3=3-m-6=-m-3<0,(x1-3)(x2-3)=x1x2-3(x1+x2)+9=m-3(3-m)+9=4m>0, that is, m>0, the value range of m is (0,1).

Variant 3 has two different signs, <==>x1x2=m<0, which is sought.

Variant 4 x1<1(x1-1)(x2-1)=x1x2-(x1+x2)+1=m-(3-m)+1=2m-2<0,==>m<1, is what is sought.

Variant 5 0

Discriminant = 0

f（1）<=0

The solution is.

First, find the axis of symmetry x=b a=-ais not in the defined domain, so the lowest point is not the axis of symmetry, so 1 is the minimum value and the positive infinity is the maximum value.

When axis of symmetry a 2<=1, f(1)<=0

When the axis of symmetry a 2>1, f(a 2) < = 0

So a>=4

You have to divide the situation, 1 there is only one root; 2 two roots;

First of all, the axis of symmetry must be in the range of x on the right side of the (x=1) axis to be between 1 and positive infinity, so the expression for the axis of symmetry is greater than 1, i.e.,

x=2 should be greater than or equal to 1, and then he can have one or two roots, so the product of a2 minus 4 times 4 is greater than or equal to 0, and the solution of the two inequalities can be combined.