Let a and B be the two real roots of the equation x2 + x-2009 = 0, then the value of A2 + 2A + B is () A. 2006B. 2007C. 2008D. 2009

Let a and B be the two real roots of the equation x2 + x-2009 = 0, then the value of A2 + 2A + B is () A. 2006B. 2007C. 2008D. 2009

∵ A is the root of the equation x2 + x-2009 = 0, ∵ A2 + a = 2009; from the relationship between the root and the coefficient, a + B = - 1, ∵ A2 + 2A + B = (A2 + a) + (a + b) = 2009-1 = 2008

2X + 3ax + A + 2A = 0 has at least one root whose module is 1

There is a root x = B + CI I is the imaginary unit B ^ 2 + C ^ 2 = 1 x, 2 (B + CI) ^ 2 + C ^ 2 = 1 x, 2 (B + CI) ^ 2 + 3A (B + CI) + A ^ 2 + 3A (b + CI) ^ 2 (B + CI) ^ 2 + 3A (b ^ 2 + 2bbci-c ^ 2) + 3A (B + CI) + 3A (B + CI) + 3A (B + CI) + 3A (B + CI I is the imaginary unit unit, B ^ 2 + C ^ 2 + C ^ 2 + C ^ 2 = 2 = 2, B ^ 2 + C ^ 2 + C ^ 2 ^ 2 ^ 2 + 2B ^ 2 + 2B ^ 2-2-2-2-2-2-2-2-2-2-2-2a + 2A + 2A + 2A + 2A + 2A + 2A + 2A + 2A + 2A + 2A + 2A + 2A + 2A + 2A + 2A + 2A + 2A + 2A + 2A + 2A + 2A + 2B ^ 2B ^ 2B ^ 2-when B = 1, 2 + 3A + A ^ 2 + 2A = 0, a ^ 2 + 5A + 2 = 0, a = (- 5 + radical 17) / 2 or a = (- 5-radical 17) / 2, B = - 1, a ^ 2-A + 2 = 0, There is no real solution. B = - A / 2 C = 1-A ^ 2 / 4 A ^ 2 + 2A + 3AB + 2B ^ 2-2c ^ 2 = 0 a ^ 2 + 2a-3a ^ 2 / 2 + A ^ 2 / 2-2 (1-A ^ 2 / 4) = 0 - 3A ^ 2 / 2 + 2a-2 = 0 3A ^ 2-4a + 4 = 0, So a = (- 5 + radical 17) / 2 or a = (- 5 - radical 17) / 2 ask: is this the case of delta greater than 0? Answer: A ^ 2 + 5A + 2 = 0, delta greater than 0 ask: is not the delta of the original equation. Minor remainder 0, a greater than 0 and less than 16 can solve a = - 1 + √ 3 answer: a root with module 1 has a complex solution, so delta greater than or less than 0 can not be used as a basis for judging whether there is a solution, 9a^2-(8a^2+16a)=a^2-16a>=0 a>=0 a

If the root of the equation 2x & # 178; + 3ax-2a = 0 about X is x = 2, then the solution of the equation y & # 178; + a = 7 about y is?

Substituting x = 2 into the original equation
8+6a-2a=0
The solution is a = - 2
y^2-2=7
y=±3

The number of soldiers in the first team is 38 less than twice that of the second team,
The number of soldiers in the first team is 38 less than twice that of the second team. If 10 soldiers are transferred from the first team to the second team, and half of the soldiers in the former second team are transferred to the first team, then the new first team has 6 more soldiers than the new second team. How many soldiers are there in the former first team and the former second team?

Suppose the original second team has x soldiers, then the original first team has (2x-38) soldiers
（2x-38)-10+x/2=x+10-x/2+6
2.5x-48=0.5x+16
2x=64
x=32
In other words, the original second team had 32 soldiers, so the original first team had (2x-38) = 2 * 32-38 = 26

Among the natural numbers from 1 to 1000, there are______ 2 or 4

There are 36 numbers from 1 to 99 that contain numbers 2 and 4; there are 36 numbers from 100 to 199, 300 to 399, 500 to 599, 600 to 699, 700 to 799, 800 to 899, 900 to 999 that contain numbers 2 and 4; there are all 2 in 100 numbers from 200 to 299, all 4 in 100 numbers from 400 to 499, and 100 numbers each containing numbers 2 and 4, so the number of natural numbers from 1 to 1000 that contain numbers 2 or 4 is: 36 × 8 + 100 × 2 = 488

The distance between city a and city B is 875 kilometers. A car leaves city a at a speed of 45 kilometers per hour. After 11 hours, how many kilometers are there from city B

875-45 * 11 = 380km

Given that angle a is more than angle B = 3:2 and angle a + angle B = 180 degrees, calculate the degree of angle A and angle B

A is three, B is two
∠a+∠B=5
180÷5=36
∠a=3×36=108
∠b=2×36=72

A car delivers goods to the mountain area, 42 kilometers per hour, 67 hours to arrive. It only takes 34 hours to return from the original road, how many kilometers per hour on average?

42 × 67 ÷ 34 = 36 × 43 = 48 (km / h). A: when returning, the average distance is 48 km / h

Given that a, B and C are any rational numbers satisfying B + C = 8 + 10A + 3A ^ 2 and c-2a-1 = a ^ 2 + 2A + 3, find the size relation of a, B and C

It's very simple to bring the second a ^ 2 = c-4a-4 into the first relation, and it becomes a simple addition and subtraction operation. Do a good job of your own

The fast and slow trains leave from a and B at the same time and meet each other after 5 hours. After meeting, the two trains continue to drive at the original speed and arrive at B after 3 hours. The slow train is 180 km away from A. how long is the highway between a and B?

The whole journey of express train takes 5 + 3 = 8 (hours). After the two trains meet, the distance of express train takes up 3 △ 8 = 38 of the whole journey, the distance of slow train takes up 38 △ 5 = 340 of the whole journey, and the distance of slow train takes up 340 × 8 = 35 of the whole journey. The highway length of a and B is 180 △ 1-35, = 180 △ 25, = 450 (km). Answer: the highway length of a and B is 450 km