Equation x square 3 2x m 0 has a real root in x 1, 1, then m is the range of values

1.Junior solution: because the coefficient before x 2 is greater than 0, the axis of symmetry = 3 4, that is, the minimum value is taken when x = 3 4, f(3 4) = -9 16-m makes f(3 4)>0 get m<-9 16, because f(-1) = 1 + 3 2-m, f(1) = -1 2-m has f(-1) > f(1).

Let f(-1)<0 get m>5 2 and get the complement as -9 16=3 4 when f-derivation》0 because f(-1)=1+3 2-m, f(1)=-1 2-m has f(-1)>f(1) let f(-1)<0 get m>5 2 take the complement as -9 16=

How do I feel like this is a bit of a problem.

Let f(x)=x 2-3 2x-m

According to the title, the axis of symmetry is x=3 4

Since x=[-1,1], 3 4 belongs to [-1,1], so f(3 4) 0 and, f(1) 0 or f(-1) 0 (with a solid root within the guaranteed range).

Solve after bringing in.

f(3 4) 0 is m 9 16

f(1) 0 is solved as m -1 2

f(-1) 0 is m 5 2

So m 9 16 and, m -1 2 or m 5 2 sum up to 9 16 m 5 2

When m=-1, the equation is a univariate linear equation and the root of the equation x=-2

When m-1, that is, the equation is a one-dimensional quadratic blind equation, because there is a real root in the collapse, (2m) 2-4(m+1)(m-3)>=0 is solved to obtain m>=3 2

In summary, the value range of m is m>=3 2 or m=-1

4 and m is not equal to 0

To sum up, there are solid roots.

m is not equal to 0, (2m+1) -4m, m is greater than or equal to -1 greater than or equal to 0m=0, and the equation is x+1=0

There is a solution. Therefore, (3 2) 2-4*1*(-m) Qinxin = 9 4 + 4m 0m J Lun -9 16

x belongs to [-1,1].

So m=x 2-3 and the first 2x, x belongs to [-1,1], and the range of m is [-9 16,5 2].

There is a question to know that m=x 2-3 2x has a root, that is, the equation has a solution, and the equation has a solution.

is to find the range of the function.

Therefore, m=x 2-3 and the width of 2x x belong to [-1,1], and the range of finding imitation m is [-9, 16, 5, 2].

Here's the answer:

First there are two real roots, satisfying = (m - 3) 4m 0 so m -10m + 9 0

m - 1）（m - 9）≥ 0

m 9 or m 1

According to Vedic theorem, there is x1 x2 = m 0, so the value of m can range from 0 m 1 or m 9

Sorry, I got it wrong, so let's fix it.

Answer: The following two conditions need to be met:

1) The equation needs to have roots, so the discriminant formula is non-negative.

Discriminant = (m-3) -4m 0

i.e. m -10m+9 0

m-1)(m-9)≥0

m 1 or m 9

2) The product of the two roots is greater than 0

Using Veda's theorem, get.

m>0In summary, the range of values of m is 0< m 1 or m 9< p >

This is a matter of the distribution of the roots.

Let f(x)=x+(m-3)x+m

m-3) 2-4m 0, f(0) = m>0 i.e. m 9 or m 1, m > 0

So m 9 or 0

1) When m=0, 3x-3=0 solves: x=1 conforms to the topic.

2) m≠0.

The swift shouting equation for x, mx -3 (m-1), x + 2m - 3 = 0, has real numbers.

i.e.: (m-3) 0

m Kuanchang Cong r

That is, the value range of m is r

Combined with the y=x-4x-m image, we get:

Its axis of symmetry x0 = 2>1

Therefore, f(0)>0, f(1) 0

i.e.: -m>0, 1-4-m 0

So, -3 m<0

Cause: 1 So when x=1 there is:

1+(m-1)+2m+6<0

3m+6<0

Solution: m<-2

Observe the image of the function f(x)=x +(m-1)x+2m+6, if there are two real roots, then the discriminant formula is 0, and f(1)<0, m -10m-23>0 and 3m+6<0

(10-root number 192) 2