Knowing that the two real roots of the equation x2 m 1 x m 1 0 are x1 x2, find the minimum value of

The two real roots of the equation x 2-(m+1)x+m+1=0 are x1, x2x1+x2=-[-m+1)]=m+1 ......x1x2=m+1 ……

(x1 2) 2 +(x2 2) 2(x1 2-4x1+4)+(x2 2-4x2+4)x1 2+x2 2+4(x1+x2)+4

x1^2+x2^2+2*x1*x2-2*x1*x2+4(x1+x2)+4

x1+x2)^2-2*x1*x2+4(x1+x2)+4 ……Put , substitute to:

m+1)^2-2(m+1)+4(m+1)+4m^2+2m+1-2m-1+4m+4+4

m^2+4m+8

m+2)^2+4

Again, make (m+2) 2+4 minimum.

m+2) 2 0 4 is the fixed value.

m+2)^2+4=0+4=4

The minimum value of x1 2) 2 +(x2 2) 2 is 4 Note: x1 2 means: x1 squared.

That's a lot, hehe.

x1+x2=m+1

x1*x2=m+1

m+1) -4 (m+1) is greater than or equal to 0

m is greater than or equal to 3 or less than or equal to -1

Original = 2 2

x1 -4x1+4+x2 -4x2+4

x1+x2) -2x1x2-8x1x2+8(m+1) -10(m+1)+8

m+1-5) -17

m-4) -17

When m=3 is the original formula there is a minimum value of -16

Solution: (x-2) + (y-2).

x2-4x+4+y2-4y+4

m-2)^2-1

m+1)^2-4(m+1)≥0

m is less than or equal to -1 or m 3

According to the image, it gets:

m=3, there is a minimum value of 0

Yes x1 2-x2 2=0 bar, sare lead x1=x2 or x1+x2=0, if x1+x2=0=2 (m+2).

m=-2 with Kai is good into the equation and the equation is unsolvable.

If x1=x2

x1+x2=2(m+2)

x1=x2=m+2

m+2)^2=2m^2-1

m^2-4m-5=0

m=5 or m=-1

Solution: baiδ=4m-4(m+3m-2)=-12m+8

du0,zhidem dao2 3。

x1+x2=-2m，x1*x2=m²+3m-2，x1(x2+x1)+x2²

x1*x2+(x1+x2)²-2x1*x2=(x1+x2)²-x1*x2

4m²-m²-3m+2

3m²-3m+2

3(m-1 2) +5 4, when inner m=1 2, the original formula is the most tolerant = 5 4.

The equation x 2-2mx+m+2=0 is solved.

x1=m-√（m^2-m-2)

x2=m+√（m^2-m-2)

Substitut equation (x1) 2+(x2) 2

(x1) 2+(x2) 2

4m^2-2m-4

The equation x 2-2mx+m+2=0 has two real roots.

So m 2 - m - 2 0

Derive m2 or m-1

So the minimum value of 4m 2-2m-4 = (2m-1) 2-5 is equal to 0

Friends on the 1st floor should consider the value range of m.

Answer: From Vedic theorem: x1 x2=2m, x1·x2=m 2, x1 x2 = x1 x2 2x1·x2= 2m 2 m 2 m 2 =4m 2m 4=4 m m 4=4[ m m 4=4 m 5, to make its value minimum, then m = 0, minimum value = 5

x1+x2=

x1-1）^2+(x2-1)^2

x1+x2)^2-2x1x2-2(x1+x2)+2=4mm-2(6+m)-4m+2

4mm-6m-10

4(m-3/4)^2-9/4-10

So the minimum value is (-49 4) when m=3 4

x1+x2=2(m+1)

x1*x2=m²-3

x1+x2)^2-(x1+x2)-12=04(m+1)^2-2(m+1)-12=0

2(m+1)^2-(m+1)-6=0

2m+2+3)(m+1-2)=0

m=-5 2 or m=1

Bring the two values back to the equation separately.

When m=-5 2, the equation is x 2+3x+13 4=0, and the verification δ=3 2-4*(13 4)<0 means that the equation has no real root, so m=-5 2 is discarded.

When m=1, the equation is x 2-4x-2=0, verify δ=4 2+4*2 0, which means that the equation has real roots, so m=1

In summary, m=1

It's x1 2-x2 2=0 x1=x2 or x1+x2=0, if x1+x2=0=2 (m+2).

m=-2 brings in the equation and the equation has no solution.

If x1=x2

x1+x2=2(m+2)

x1=x2=m+2

m+2)^2=2m^2-1

m^2-4m-5=0

m=5 or m=-1

According to Vinda's theorem, there is x1+x2=1 2(m-1), x1*x2=1 2(m+1), and because x1-x2=1

So 1=(x1-x2) 2=(x1+x2) 2-4x1x2=[1 2(m-1)] 2-4*1 2(m+1), which is simplified to m 2-10m-11=0

m-11)(m+1)=0

i.e. m=11, or m=-1

When m=11, the original equation is: 2x 2-10x+12=0x 2-5x+6=0

x-2)(x-3)=0

x = 2 or x = 3

When m=-1, the original equation is: 2x 2+2x=0x(x+1)=0

x=-1 or x=0

Answer: When m=-1, the root of the original equation is -1, and when m=11, the root of the original equation is 2, 3

Or. According to the relationship between the root remainder collapse coefficient :

x1+x2=(m-1)/2---1)

x1*x2=(m+1)/2---2)

and x1-x2=1 --3).

Just solve it

From the title: m+1 is not equal to 0, that is, m is not equal to -1

Let y=f(x) =(m+1)x 2+2(2m+1)x+1-3m, when m+1<0, i.e., m<-1: x1<10 f(3)=9(m+1)+6(2m+1)+1-3m=18m+16<0

Solution: -20, i.e., m>-1: f(1)=m+1+2(2m+1)+1-3m=2m+4<0 f(3)=9(m+1)+6(2m+1)+1-3m=18m+16>0

Solution: No solution.

The value range of m is -2