It is known that the equation about x x 2 2 m x 3 6m 0

2、（2003?Yancheng) is known to have the equation for x2+2(2-m) x+3-6m=0

1) Verification: No matter what real number m takes, the equation always has a real number root;

2) If the two real roots x1 and x2 of the equation satisfy x1=3x2, find the value of the real number m Test point: the discriminant formula of the root; Solving a quadratic equation - factorization; The Relationship Between Roots and Coefficients Special Topic: Calculation Problems; Proof Analysis:

1) Prove that the discriminant formula of the root of a quadratic equation is everstable at 0, which can be solved;

2) According to the relationship between the root and the coefficient of the unary quadratic equation x1 + x2 = 4x2 = -2 (2-m) = 2 m-4, and x1?x2=3x22=3-6m to find the value of m Answer: Solution:

1) Prove: In the equation x2+2(2-m)x+3-6m=0, =4(2-m)2-4(3-6m)=4(m+1)2 0, no matter what real number m takes, the equation always has a real root

2) If the two real roots of the equation x1 and x2 satisfy x1 = 3x2, then x1 + x2 = 4x2 = -2 (2-m) = 2 m-4

x2= m2-1 ①

x1?x2=3x22=3-6m, x22=1-2m, substitute m(m+4)=0, that is, m=0, or m=-4

Answer: The value of the real number m is 0 or -4 Comment: The key to solving this question is to be familiar with the relationship between the case and the discriminant of the root of a quadratic equation, and the relationship between the root and the coefficient

1）△＞0?The equation has two unequal real roots;

2）△=0?The equation has two equal real roots;

3）△＜0?Equations do not have real roots

4) If a quadratic equation has a real root, then x1+x2=- ba, x1x2= ca

Look closely.

Equations are factored in.

x+3)[x+(1-2m)]=0

Its two roots are -3 and 2m-1

When x1=-3, then x2=2m-1

x1=3x2

2m-1=-1

m=0 When x2=-3, then x1=2m-1

x1=3x2

2m-1=-9

m=-4 In summary, the value of m that satisfies the condition is 0 or -4

Discriminant = (m+2) -4(2m-1).

m²+4m+4-8m+4

m²-4m+4+4

m-2)²+4≥4>0

So the equation has two unequal real roots.

The opposite number is x1+x2=0

Vedic theorem. x1+x2=-(m+2)=0

m=-2 is now x -5=0

x²=5x=-√5,x=√5

1) Equations have two real roots.

=4(m+1)²-4m²≥0

That is: 8m+4 0

Solution: m -1 2

2) When m>-1 2, δ>0 the original equation has two unequal real roots, so m=0 can be taken

In this case, x -2x=0 solves: x=0 or 2

If there are two real roots, there is:

Discriminant = 4 (m+1) 2-4m 2> = 0

Then 4m 2 + 4 + 8m - 4m 2 > = 0

then m>=-1 2

Based on the results above, ...m=-1 2, the equation has two equal real roots, then only the integer of the region -1 2 is sufficient.

For example, take m=0

The equation is x 2-2x = 0

x=0 or 2 to meet the conditions.

1) Equations have two real roots.

4(m+1)²-4m²≥0

Solution: m -1 2

2) When m>-1 2, δ>0 the original equation has two unequal real roots, so m=1 can be taken

(1) m is greater than -1 2 (2) is a lot, meet (1) to find a suitable one.

1. That is, the discriminant formula is greater than or equal to 0

So [-2(m+1)] 4m 0

4m²+4m+1-4m²≥0

m≥-1/4

2. Let m=0

then x -2x = 0

x(x-2)=0

x=0,x=2

x1=3x2

x1=ax2=b

Then: x1+x2=4x2,x1+x2=-2(2-m)x2=(m-2) 2

x1x2=3x2^2,x1x2=3-6m

x2^2=(3-6m)/3=1-2m

So, ((m-2) 2) 2=1-2m

m^2-4m+4=4-8m

m^2+4m=0

m = 0 or -4

The value of the real number m is 0, or, -4

x1=a

x2=b then: (root and coefficient relationship).

x1+x2=-2(2-m)

x2=(m-2)/2

x1x2=3-6m

x2^2=(3-6m)/3=1-2m

So, ((m-2) 2) 2=1-2m

m^2-4m+4=4-8m

m^2+4m=0

m = 0 or -4

Using the root finding formula directly, the two roots are 2m-1 and -3 respectively, so 2m-1 = -9 or -1, m=-4 or 0

In the case of factorization, it is (x-2m+1)*(x+3)=0, and the two are still 2m-1 and -3

a+b=-2(2-m) ①

a×b=3-6m ②

a=3b ③

Substituting the formula into the formula respectively, we get: b= (m-2) b =1-2m

Combine the two formulas to obtain: m +4m=0

So, m = 0 or -4

1) The root number in the solution of the square orange is (m+2) -4(2m-1)=(m-2) +4>0, so the equation has two real roots of unequal bent cultures.

2) Bring 1 in and get m = 2

There is another root that is x=4

So the circumference of the triangle is 1+4 + root number 17 or 1+4 + root number 15

Little Fish:

What your title means is not the sum of squares of two roots.

Let the two roots be x1 x2

Then there is: x1+x2

2（m-2）

2m-4x1*x2=m²

x1²+x2²

x1+x2）²-2x1x2

2m-4）²-2m²

4m²-16m+16-2m²

2m²-16m+16

According to the title. 2m²-16m+16=56

2m²-16m-40=0

m²-8m-20=0

m+2）（m-10）=0

m= -2 m=10

Let's consider the range of m values.

b²-4ac

4（m-2）²-4m²

16m+16

0-16m+16≥0

m 1 only m=-2 satisfies the topic.

In summary, m= -2

The equation has two real roots, discriminant 0

4(m-2)^2-4m^2≥0

m-1 0m 1 is set to two real roots, which are x1 and x2

x1^2+x2^2

x1+x2)^2-2x1x2

4(m-2)^2-2m^2

4m^2-16m+16-2m^2=56

m^2-8m-20=0

m-10)(m+2)=0

m = 10 or m = -2

m 1 again, so m = -2

Assuming that there are roots, let the two roots be x1 and x2

then x1+x2=2(m-2).

x1x2=m²

Then x1 +x2 =(x1+x2) -2x1x2=4(m-2) -2m =2m -16m+16=56

Collating the equation yields m -8m-20, i.e., m = 10 or -2 and 4 (m-2) -4m 0, i.e. m 1

So m=-2

=(m+2) 2-4(2m-1)=m 2-4m+8=(m-2) 2+4 Evergrande is 0

Therefore there must be two unequal real roots.

By the Vedic theorem.

x1+x2=-(m+2)

And x1 and x2 are inverse numbers to each other, so x1+x2=0 so m+2=0

m=-2 becomes the original equation.

x^2-5=0

x= root number 5

=（m+2）²-4（2m-1）=m²-4m+8=（m-2）²+4＞0；Therefore there must be two unequal real roots.

If they are opposites, then x1+x2=-b 2a=-(m+2) 2=0;m=-2 can be obtained, and the original equation is x -5=0 to obtain x= root number 5

x²-(m+2)x+(2m-1)=0

-(m+2))²4(2m-1)

m²+4m+4-8m+4

m²-4m+4+4

m-2)²+4

The equation has two unequal real roots;

1²-(m+2)*1+(2m-1)=0

m-2=0m=2

x²-4x+3=0

x-1)(x-3)=0

x1=1x2=3

Another root of the equation: x=3

is a right-angled side, and the hypotenuse is long: (1 +3 )= 10The circumference of a right-angled triangle: 1+3+ 10=4+10

Discriminant formula by root: (m+2) -4 1 (2m-1)=m +4m+4-8m+4=(m-2) +4>0

So the equation has two unequal real roots.

The two are opposites, i.e. m+2=0 and m=-2

Solution: Since sin ,cos are the two roots of the equation, there is sin +cos =6m (1).

sinα*cosα=2m+1 （2）

Square (1) to get the following

1+2sin *cos = 36m, substituting (2) gets:

1+2(2m+1)=36m²

1+4m+2=36m²

36m²-4m-3=0

to solve m

From the properties of the equation, we know that x1+x2=-b a x1*x2=c a, so sina+cosa=-6m 3=-2msina*cosa=(2m+1) 3

By sina +cosa =1 (sina + cosa) =sina +cosa +2*sina *cosa = 1 + 2 * (2 m + 1) 3 = 4m

The solution yields m=-1 2 or m=6 5