When m is what the value, the equation x 2m 2 x m 5 0

This is mainly based on the discriminant formula of the root.

Knowledge preparation: For a unary quadratic equation ax +bx+c=0, the discriminant of the root =b -4ac, when 0 the equation has 2 unequal real roots, when =0, the equation has two equal real roots, sometimes say one, when 0, the equation has no root.

In this problem, =(2m+2) -4*1*(m +5)=8m-16 When 0, i.e., m 2, there are two unequal roots.

When =0, i.e., m=2, there are two equal roots.

When 0, i.e., m2, there is no root.

Do you understand that?

Discriminant formula of unary quadratic function: =(2m+2) 2-4(m 2+5)=8m-16

There must be two unequal real roots, and the discriminant formula is greater than 0, that is, 8m-16>0, so m>2 has two equal real roots, then the discriminant formula is equal to 0, that is, 8m-16=0, so m=2 has no real roots, then the discriminant formula is less than 0, that is, 8m-16<0, so m<2

Finding >0 has two unequal real roots.

0 does not have a real root.

There are two equal real roots.

Summary. When m=1, the equation x +2x+3=0

What to do with the equation x + (3m-1) x + 2m -1 = 0 when m is what is the value. When m=1, the equation x +2x+3=0 continues. Isn't this okay?

Wrong. Wrong. Process? Good.

Please tell me about the process.

When m is an arbitrary real number, the solution of the stupid number x + (3m-1)x+2m -1=0 is: x = 3m-1 9m -4(2m -1))] 2= [3m-1 9m -8m +4)] 2= [3m-1 Zhongxiang (m +4)] 2= [3m-1 m+2)] 2= (3m-1+m+2) 2 or (3m-1-m-2) 2= 2m+1 or m-3

When m is what the value, the equation x + (3m-1) x + 2m -1 = 0 has a real root.

When m=1, the equation x + (3m-1) x + 2m -1 = 0 has a real root. Solution: Let the equation x + (3m-1)x+2m sell the two real roots of -1=0 as x1 and x2, then there are x1+x2=-(3m-1), x1x2=2m with pick-1, x1=-1+2m, x2=-1-2m, since x1 and x2 are both real numbers, so when m=1, the equation x + (3m-1)x+2m -1=0 has real roots.

=(2m+2)^2-4(m^2+5)

4m^2+8m+4-4m^2-20

8m-161) When 8m-16>0, i.e., m>2, there are two unequal real roots.

Yu Zhi 2) When 8m-16=0, that is, m=2, there are two real roots of the same number as the coincidental number.

3) When 8m-16<0, the key is destroyed.

That is, m<2, there is no real root.

a（1+x²)+2bx-c(1-x²)=0a+c)x^2+2bx+a-c=0

4b 2-4(a+c)(a-c)=4b2-4a2+4c2=0, so b2+c2=a2

So the triangle ABC is a right triangle and A is a right angle.

The equation for x (m+1) x + (1-2x) m = 2, what is the value of m, first sort out: (m+1)x -2mx + m - 2 = 0= b -4ac = 4m -4(m+1) (m - 2) =4(m+2).

1) The equation has two unequal real roots.

4(m+2) >0, i.e., m > 2, there are two unequal real roots.

2) The equation has two equal real roots.

4(m+2) =0 , i.e. when m = 2, there is a second-order solid root.

3) The equation has no real roots.

4(m+2) <0, i.e., m < 2, there is no real root.

4) Equations have real roots.

m 2, there are real roots.

The equation for x (m+1) x + (1-2x) m=2 is deformed to be: (m+1)x -2mx+m-2=0

When the equation has two unequal real numbers followed by:

b 2-4ac>0, i.e. =(2m) 2-4*(m+1)*(m-2)>0

Simplification: m>-2;

When the equation has two equal real numbers followed by:

b 2-4ac>0, i.e. =(2m) 2-4*(m+1)*(m-2)>0

Simplification: m>-2;

When the equation is not followed by real numbers:

b 2-4ac>0, i.e. =(2m) 2-4*(m+1)*(m-2)>0

Simplification: m>-2;

4) Equations have real numbers to follow.

b 2-4ac>0, i.e. =(2m) 2-4*(m+1)*(m-2)>0

Simplification: m>-2;

m+1)x²-2mx+m-2=0

-2m) 2-4(m+1)(m-2)=4m+81) 0 so m -2

2) =0 so m=-2

3) 0 so m -2

4) 0 so m -2

Because 5*m*m-4*(m+2)*(m-3)=(m+>0), the equation has real roots.

Let a=(m+;

x1=(√5m+√a)/(2*(m+2));x2=(√5m-√a)/(2*(m+2));

by x1+x2=3;

It can be seen that: 5m (m+2)=3;

Good to God: 5m 3*(m+2);

Launch: (5-3)*m=6;

Answer: m=6 ( Traveler Pb 5-3);

Hope it helps.

Have you ever learned about root control?

Question 1: If there are two positive real roots, then the equation must be a quadratic function, so the quadratic coefficient m+1 is not equal to 0, and m is not equal to -1

Discriminant formula = b 2-4ac=4(4m 2+4m+1)-4(m+1)(1-3m)>=0, (guaranteed to have 2 real roots).

4m^2+4m+1+3m^2+2m-1>=0

7m^2+6m>=0

m<=-6 7 or m>=0

The axis of symmetry x=-b (2a)=-(2m+1) (m+1)>0, (condition 1 of the two positive roots).

10, (condition 2 of the two positive roots).

10), the y-axis intercept 1-3m is less than 0, when the opening is downward (m+1<0), the y-axis intercept 1-3m is greater than 0, so:

m+1)(1-3m)<0

m<-1 or m>1 3

So m is in the range of (negative infinity, -1) and (1 3, positive infinity).

The equation has two real roots.

4(2m＋1)²－4(m＋1)(1－3m)＞0 ∴7m²＋6m＞0 ∴m(7m＋6)＞0

m 0 or m 6 7

Let the two roots be x1 and x2, then x1 x2 2 (2m 1) (m 1) x1x1 (1 3m) (m 1).

1) There are two solid roots x1 x2 0 and x1x2 0

2 (2m 1) (m 1) 0 and x1x1 (1 3m) (m 1) 0 1 m 1 2 and 1 m 1 3

1＜m＜﹣1/2

m 0 or m 6 7 1 m 6 7

2) There is a positive and a negative two solid roots x1x2 0

1 3m) (m 1) 0 m 1 or m 1 3 m 0 or m 6 7 m 1 or m 1 3

(1) An equation has two equal real roots.

=（4 m）² 4 × 2（2 m ² m）= 016 m ² 16 m ² 8 m = 08 m =0

m = 02) The equation has two real roots.

= 16 m ² 16 m ² 8 m ≥ 0∴ 8 m ≥ 0

m ≥ 0

Knowing the equation for x: 2x +4mx+2m -m=0, when m is the value, the equation 1There are two equal real roots.

16m² -8(2m²-m)

16m²- 16m²+ 8m

8m = 0

m=02.There are two real roots.

8m > 0

m > 0

When b -4ac = (4m) -4 2 (2m -m) = 0, i.e., m=0, the equation has two equal real roots;

When b -4ac = (4m) -4 2 (2m -m) > = 0, i.e., m>=0, the equation has two real roots.

△=4m²-4m-8

1) If one of the two roots of the equation is greater than 0 and less than 0, it must be >0 and m+2<0

So m<-2

2) If both roots of the equation are positive, must.

0 and m+2>0, -2m>0

The solution is -20 and (x1-1) (x2-1) <0, i.e., m+2+2m+1<0 is m<-1

If there are two roots of the equation, there is:

4m^2-4(m+2)>0

m^2-m-2>0

Solution: m>2 or m<-1

1) Why is the value of m, one of the two roots of the equation greater than 0 and less than 0?

Then there is: x1x2<0

Get: m+2<0 i.e.: m<-2

2) Why is m a value when both roots of the equation are positive?

Then there are: x1x2>0, and x1+x2>0

Get: m+2>0

2m>0

In summary, it is found that when -2(3)m is valued, one of the two roots of the equation is greater than 1 and less than 1?

Then there are: (x1-1)(x2-1)<0

x1x2-(x1+x2)+1<0

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

m+2m+3<0

m<-1

1x square.

2x square. 3x square.

4m²-4m-8

1) If one of the two roots of the equation is greater than 0 and less than 0, it must be >0 and m+2<0

So m<-2

2) If both roots of the equation are positive, must.

0 and m+2>0, -2m>0

The solution is -20 and (x1-1) (x2-1) <0, i.e., m+2+2m+1<0 is m<-1