Binary Equations, Rush and Hurry!! If it s good, add points, be sure to do it within 1 hour!

The discriminant formula for x -(2k+1)x+4(k- 1 2) is:

2k-1)² 4×4(k - 1/2)

4k² +4k + 1 - 16k + 84k² -12k + 9

2k-3)²

So when k = 3 2, the discriminant formula is 0, and there are two equal real roots;

When k is not equal to 3 2, the discriminant formula is always positive, and there are two unequal real roots.

The two roots of the equation are.

2k+1) +2k-3)) 2 = 2k - 1 and. (2k+1) -2k-3)) 2 = 2, so the length of the three sides of the triangle is 4,2,2k-1

Because this is an isosceles triangle, 4 = 2k-1 or 2=2k-1 so k = 5 2 or k = 3 2

That is, the three sides of the triangle are 4,4,2 or 4,2,2, respectively, and the latter needs to be rounded off because the sum of the two sides of the triangle is greater than the third side.

That is, when k = 5 2, the length of the three sides is 4,4,2

1)b²-4ac

2k+1)²-4*4(k-½）

4k²+4k+1-16k+8

4k²-12k+9

4（x-3/2）²

This equation is greater than or equal to 0, so there are two roots or one root.

2) Incomplete.

1、a=1，b=-2k-1，c=4k-2

b^2-4ac=(2k+1)²-4*1*(4k-2)4k²+4k+1-16k+8

4k²-12k+9

Match him to a perfectly flat pattern: 4 (x-3 2).

No matter what plant x takes, this perfectly flat method is always greater than or equal to 0, so there are two unequal real roots or two equal real roots.

2 If this equation is the above equation, then discuss it in different situations:

The waist of isosceles is 4: according to Veda's theorem (relationship with coefficients): x1+x2=-b a, x1*x2=c a

Let the bottom edge be x:4+x=2k+1, and 4x=4k-2 will be solved to get k=5 2, x=2

If the base of the isosceles is 4, then let the waist length be y (it is better not to set two identical unknowns in a problem), then y+y=2k+1, y=4k-2

The solution yields k = 3 2 and y = 2

Because it does not conform to the trilateral relationship, y=2 is rounded.

So the side length of the isosceles triangle is

1.Factor the left side of 6x 2-5xy+y 2=0 to get (2x-y)(3x-y)=0

2x-y=0 or 3x-y=0, y=2x or y=3x

When y=2x, 2x=x 2+6x+4, x 2+4x+4=0, (x+2) 2=0, x=-2, y=2x=-4

When y=3x, 3x=x 2+6x+4, x 2+3x+4=0, the equation has no solution.

The original system of equations has a solution: x=-2, y=-4

2.Direct substitution.

Substituting y=mx+2 into the following equation gives (mx+2) 2+4x+1=2mx+4

m^2x^2+4mx+4+4x+1=2mx+4,m^2x^2+(2m+4)x+1=0

(2m+4)^2-4m^2=4m^2+16m+16-4m^2=16m+16

16m+16<0,16m<16,m<-1

3.△=4k+1)^2-8(2k^2-1)=16k^2+8k+1-16k^2+8k=16k+1

16k+1>0,16k>-1,k>-1/16

4.Use "Relationship between Root and Coefficient".

m+n=2m-1,n=m-1

mn=m^2+1,m(m-1)=m^2+1,m^2-m=m^2+1,m=-1

n=m-1=-2

m n = (-1) (2) = 1 (-1 to the even power of 1).

It seems that you are not in a hurry, this is all offline)

x^2-2/(m-1)x+3/(m-1)=0x1*x2=3/(m-1)

When m-1 0, i.e., m1, x1*x2 0, i.e., there is a positive root and a negative root x1+x2=2 (m-1) 0

So the absolute value of the negative root is large.

(m-1)x-2x+3=0 has a positive and negative root, x1x2=3 (m-1)<0

m<1m-1<0

x1+x2=2/(m-1)<0

So the absolute value of the negative root is large.

Let y=(m-1)x·x-2x+3, the two roots of the equation are the two intersections of the function and the x-axis.

When x=0, y=3, that is, the function intersects the y-axis at the point (0,3), if you want to make the equation have two different signs, then m-1<0,m<1, the symmetry axis of the function is x=1 (m-1)<0, you can draw a rough picture of the function from the above information, and it is obvious that the absolute value of the negative root is large.

If you want to write a process, you can solve it according to the sum of the two roots (less than 0).

1, first of all, there are two roots of this equation, so the equation is written in the form of ax*x+b*x+c=0, and then b*b-4ac>0, the formula applied to this equation should be (-2)*(2)-4*(m-1)*3>0 to solve m<4 3

2, and because the equation has a positive root and a negative root, the product of the two roots (c a) < 0, that is, 3 (m-1)<0 is solved to obtain m<1

In summary, m<1

To see which of the two is the absolute value of the two, it depends on whether the sum of the two is positive or negative, and the sum of the two is (m-1) 2

Because m<1, (m-1) 2<0So the absolute value of the negative root is greater than the absolute value of the positive root.

The simultaneous equations are used to solve the troubles, and the detailed process is described in the lead bend.

Solution: Let the velocity of A be xkm h and B be ykm h

4x+150=4y

Solution x=y=

Suppose A's velocity is x, and B's velocity is y

1) If two cars start at the same time and go in the same direction, car B can catch up with car A in 4 hours. Equation 4y=4x+150, so y-x=35

2) Go in the opposite direction, and the two cars meet in an hour. So x+y=100, solve the system of equations x=65 2=, y=35+

So the speed of car A is, and the speed of car B is.

Solution: Let the average speed of cars A and B be x kilometers and y kilometers respectively, according to the question, 4 (y-x) = 150

Solving the system of equations yields, x=, y=A:

Suppose the speed of vehicles A and B is x y kilometers per hour.

4y=4x+150

Let the speed of A and B be x y

then 150 x + 4 = 150 y

The solution is 75 25

Untie; Let the speed of A and B be x kilometers and y kilometers.

Just put this one in again

The most annoying thing about the Olympiad Horse!!

1、 x^2-3x-1=0

x-3/2)^2-9/4-1=0

x-3/2)^2=13/4

x-3 2 = root number 13 2 or x-3 2 = - root number 13 2x = (3 + root number 13) 2 or x = (touch eggplant 3 - root number 13) 22, x 2-7x + 11 = 0

x-7/2)^2-49/4+11=0

x-7/2)^2=5/4

x-7 2 = root number 5 2 or refers to Zheng x-7 2 = - root number 5 2x = (7 + root number 5) 2 or x = (7 - root number 5) 23, 5x 2 = 2x+1

5x^2-2x-1=0

x^2-2x/5 -1/5=0

x-1/5)^2-1/25-1/5=0

x-1/5)^2=6/25

x-1 5 = root number 6 5 or x-1 5 = - root number 6 5x=(1 + root number 6) 5 or x=(1-root number 6) 5

a(1-x)-2*root2)*bx+c(1+x)=0 is organized into (c-a)x -(2*root2)*bx+(a+c)=0 with a solution, then there is [(2*root: 2)*b] -4(c-a)(a+c)>=0 to a +b -c >=0(1).

And by a, b, c are the three sides of the RT triangle abc, c=90, there is a + b -c =0, so the inequality (1) takes the equal sign, so the equation (c-a) x -(2 * root number 2) * bx + (a + c) = 0 has two equal real roots, i.e. x1 = x2

From x1+x2=12, we can get x1=x2=6, from Vedic theorem to get two equations, and then substitute x=6 into the equation (c-a)x-(2*root number 2)*bx+(a+c)=0 to get the third equation, and solve the above three equations about a, b, and c to get a, b, c

I found the value of x.

x = 6 3 times root number 2

I think this is calculated, and the rest should be easy to calculate, and the other = b -4ac = 8b -4 (c -a) = 4b to get 2b with the root number

Therefore x = (b*root2 b) (c - a).

3(x-1)(x-m)-(7-m 2)x=03x 2-3(m+1)x-3m-(7-m 2)x=03x 2-(3m+3+7-m 2)x-3m=03x 2+(m 2-3m-10)x-3m=0, the two real roots are inverse.

First, there are two real roots, and the discriminant formula is greater than or equal to 0

m^2-3m-10)^2+4*3*3m≥0[(m+2)(m-5)]^2+36m≥0………1) Two are opposite numbers to each other, according to Vedd's theorem.

m^2-3m-10)=0

m^2-3m-10=0

m+2)(m-5)=0

m = -2 or m = 5

m=-2, (1) is not true.

So m=5

Please write clearly, square is very different from x, and besides, the multiplication sign and x look very similar!

m=-2

The calculation is as follows: the equation can be arranged as 3x·x+(m·m-3m-10)x+3m=0 If two real numbers are opposite to each other, then the coefficient of the primary term is 0, i.e., m·m-3m-10=0, (m+2)(m-5)=0

We get m1=-2 , m2=5

where m2=5 is not a real root, which does not fit the meaning of the question, so m=-2