A master solves a problem of a quadratic equation with a trinity

mx2-4x+5=0 (1)

x2-4mx+4m2-3m-3=0 (2) can be obtained from (1)-(2).

m-1)x2+(4m-4)x-4m2+3m+8=0 because when m=1, there is no solution.

So m is not equal to 1

Because systems of equations have real roots.

So. = (4m-4)2-4(m-1)(-4m2+3m+8)16(m-1)2-4(m-1)(-4m2+3m+8)(m-1)[16(m-1)-4(-4m2+3m+8)]4(m-1)(4m2+m-12) is greater than or equal to zero.

When m is greater than 1, 4m2+m-12 is greater than or equal to 0

When m is less than 1, 4m2+m-12 is less than or equal to 0

First, both equations have roots, which means: (-4) 2-4*m*5>=0; (a 2 represents the square of a, > = means greater than or equal to) simplification to m<=; (-4m) 2-4(4m 2-3m-3)>=0, simplified to m(4m 2-m+3)<=0, we can know m<=0; m is not equal to 0; So m<0; As long as m<0, both equations have roots, here assuming m=-1; Then the heels of the first equation are 1 and -5.

The second equation is followed by -2; Hence the existence;

There are two equal real roots.

The discriminant formula is 0b -4a = 0

b = 4a, so the original formula = a*4a [(a-2) +4a-4]=4a (a -4a + 4 + 4a-4).

4a²/a²=4

If there are two equal real roots, b-square-4ac, that is, b-square-4a equals 0, b-squares equals 4a, the original formula equals 4a, a2-4a+4+4a-4 equals 4a, square a, square a, square 4 equals 4a, square a-square equals 4

1. From the equation (x-3) (x-6) = 0 to get x=3 or x=6 Since the other two sides of the triangle are 4 and 7, the third side should be 6

2. From the equation x -2x-48 = 0 to get (x + 6) (x - 8) = 0 to get x = -6 or x = 8 so x1 = -6 and x2 = 8 are obtained

3. Left-Right =(A 2-12A+36)+(B 2-16B+64)+(C 2-20C+100) =(A-6) 2+(B-8) 2+(C-10) 2=0 Therefore, A=6, B=8, C=10 by A 2+B 2=100=C 2 A triangle is a right-angled triangle with C as the hypotenuse and A, B as the right-angled side.

1. Solve the equation: x1 = 6, x2 = 3 (rounded off, because 4 + 3 = 7) perimeter: 4 + 7 + 6 = 17.

2. x square-2x=48, x-1=7, x1=-6 (rounded), x2=8, the square of this number is 64.

3. Original formula: a(12-a)+b(16-b)+c(20-c)=2000a<12,b<16,c<20, and a=6, b=8, c=10, the maximum value on the right side of the equation is: 200.

It can't be equal to 2000. Wrong question! It should be 200.

In this case, the triangle ABC is a right triangle.

2 (x-2) squared = 4-x squared.

2 (x square - 4x + 4) - 4 + x squared.

2x square + 8-8x-4 + x squared.

3x square - 8x+4

x-2)(3x-2)

x1=2 x2=2/3

(1) Vedder theorem: b 2-4ac = (2k-3) 2-4k 2 = 9-12k> = 0

So k<=3 4

2) =6 i.e. 3-2k+k =6 solution: k=3 (rounded) or k=-1

So the original equation becomes: x -5x + 1 = 0

Upstairs is correct, no problem. The idea is correct, the calculation is not seen, and it should be correct.

1 △=(2k-3)²-4k²=-12k+9>0k<3/4

2 α+= -(2k-3),β= k²

+= -(2k-3)+k = 6 The solution is: k = -1 or 3

From (1) k<3 4

So: k=-1, x -5x+1=0

- 3 -5 = 21+3 1-5 = 19I'm also in the third year of junior high school

(1) Because there are two real roots.

So =b -4ac=(2k-3) -4k =-12k+9>03 i.e. k<-4b

2) Because + =--=-(2k-3), c a

=-=k a + = -(2k-3)+k = 6 The solution is: k = -1 or k = 3

3 because k<-

4 so k = -1

then k=-1, x -5x+1=0

1, x + y -5 = 2 or -2, so x + y = 3 or 72, x = 1 substitution, a + b + c = 0 --1 formula x = -2 substitution, 4a - 2b + c = 0 --2 formula 1 and the left and right sides multiply by 2, 2a + 2b + 2c = 0 --3 formula 2 formula, 3 formula, the left and left sides are added, the right and right sides are added to get 6a + 3c = 0, and the left and right sides are divided by 3, 2a + c = 0

If (x + y -5) = 4, then x + y = 3 or 7

If the two roots of the equation ax + bx + c = 0 (a is not equal to 0) for x are 1 and -2 respectively, then 2a + c = 0