How to calculate (50-40 + x) (500-10x) = 8000

How to calculate (50-40 + x) (500-10x) = 8000

（50-40+X）（500-10X）=8000
-10 (-30 + x) (-10 + x) == 0
x=30
x=10
（50-40+X）（500-10X）=8000 10
How to solve (x-40) (500-10x) = 8000
Hurry, speed!
500x-20000-10xx+400x=8000
-10xx+900x=28000
xx-90x+2800=0
(x-45)(x-45)=-775
unsolvable
Simplification: (x-40) (x-50) = - 800
X^2-90X+2800=0
Use the root formula △ = 90 ^ 2-4 * 2800 = - 3100
So there is no real root
The following formula of quadratic equation of two variables
x*x-90x+1000=0
Now let's solve it with a quadratic equation of two variables.
No, I learned it myself
（x-40)(500-10x)=8000
500x-10x²-20000+400x=8000
-10x²+900x-20000=8000
-10x²+900x-12000=0
But ∵ B & sup2; - 4ac = - 3100 < 0
The equation has no solution
Two elementary three one variable quadratic equations application problem. Now will
1. There is a two digit number. The number on each digit is 5 larger than the number on the tenth digit. If two numbers are exchanged, the sum of the new number and the original number is 143. Find two numbers
2. Someone saves 800 yuan in two forms, one with an annual interest rate of 10% and the other with an annual interest rate of 11%. He gets a total interest of 85 yuan and 50 Jiao. How much money has he saved in the two forms
1. The original digit of this number is x, so the number on the tenth digit is X-5 10x + (X-5) + 10 × (X-5) + x = 143
10X+X-5+10X-50+X=143
22X-55=143
22X=143+55
22X=198
X=198÷22
X=9
So the number ten is 9-5 = 4
So this double-digit number is 49
2. If one kind of deposit is x yuan, the other is (800-x) yuan
10%X+（800-X）×11%=85.5
10%X+88-11%X=85.5
1%X=2.5
X=2.5÷1%
X=250 800-250=550
A: one 250, the other 550
The quadratic equation of one variable is solved by formula method
X ^ 2 - (2 + radical 2) x + radical 2-3 = 0
x^2-(2+√2)x+√2-3=0
△=(2+√2)^2-4(√2-3）
=6+4√2-4√2+12
=18
√△=3√2
x1=（2+√2+3√2）÷2=1+2√2
x2=（2+√2-3√2）÷2=1-√2
Can we find the value of X + 2Y + 3Z by solving the binary linear equation under the condition of 2x + y + 3Z = 23, x + 4Y + 5Z = 36?
2x+y+3z=23
x+4y+5z=36
If x is eliminated, the following results are obtained
7Y＋7Z＝49
So y + Z = 7
So y = 7-z
from
2x+y+3z=23
x+4y+5z=36
If y is eliminated, the following results are obtained
7X＋7Z＝56
So x + Z = 8
So x = 8-z
So x + 2Y + 3Z = (8-z) + 2 (7-z) + 3Z = 22
If the two roots of the univariate quadratic equation AX ^ - 2x-1 = 0 have the same sign, then the value range of a is
If the two roots of the quadratic equation AX ^ - 2x-1 = 0 of X have the same sign, then the value range of a is?
To simplify the process, which expert to guide?
Two roots of equation AX ^ - 2x-1 = 0 have the same sign
x1x2=-1/a>0
That is: a = 0
That is: a > - 1
To sum up, - 1
There are the following propositions: (1) if 3x + 2Y = 0, then x = y = 0; (2) if x (1-x) = 0, then x = 0; (3) the quadratic equation AX2 + BX + C = 0, if AC < 0, then the equation must have a real number solution; (4) if (x − 1) 2 = x − 1, then x > 1, where is the true proposition______ .
① If 3x + 2Y = 0, we can get 2Y = - 3x, we can get many results, so x = y = 0 is a false proposition; if x (1-x) = 0, we can get x = 0, or x = 1, so ② is a false proposition; in a quadratic equation of one variable, the discriminant b2-4ac > 0, then the equation has two unequal real roots, because AC < 0, then the discriminant b2-4
If the function f (x) = 4x2-kx-8 is monotone on [5,8], then the value range of K is______ .
According to the properties of quadratic function, we know that the axis of symmetry x = K8 is a monotone function in [5,8], then the axis of symmetry can not be in this interval 〈 K8 ≤ 5, or K8 ≥ 8, K ≤ 40, or K ≥ 64. So the answer is: (- ∞, 40] ∪ [64, + ∞)
Solve the equations 3x-y + 2x = 20 2x + Y-Z = 40 7x + y + 5Z = 55
(1) 3x-y + 2Z = 20 (should 2x of the latter be 2Z)
(2) 2X + Y-Z = 40 (add with (1) to get (4)
（4）5x+z=60
(3) We get (2 + 5) by subtracting (2 + 5)
(5) 5x + 6Z = 15 (minus (4))
5z=-45
Z = - 9 (substitute to (4)
5x=69
x=69/5
Substituting x = 69 / 5; Z = - 9 into (2)
y=40-2x+z=40-2(69/5)+(-9)=17/5
checking calculation:
(1) Left = 3 (69 / 5) - (17 / 5) + 2 (- 9) = (207-17-90) / 5 = 100 / 5 = 20
(2) Left = 2 (69 / 5) + 17 / 5 - (- 9) = (138 + 17 + 45) / 5 = 200 / 5 = 40
(3) Left = 7 (69 / 5) + 17 / 5 + 5 (- 9) = (483 + 17-225) / 5 = 275 / 5 = 55
Therefore, the original equations hold
X=23/65 Y=785/23 Z=299/23
If the equation AX2 + 2x + 1 = 0 about X has only negative real roots, then the value range of real number a is______ .
(1) When a = 0, the equation is a linear equation with a negative real root (2) when a ≠ 0, when the equation AX2 + 2x + 1 = 0 has a real root, △ ≥ 0, the solution can be a ≤ 1; when the equation AX2 + 2x + 1 = 0 has a negative real root, 1A < 0, the solution can be a < 0; when the equation AX2 + 2x + 1 = 0 has two