Problems related to quadratic equation of one variable When a=_____ ,b=_______ The equation x ^ 2 + 2 (1 + a) x + (3a ^ 2 + 4AB + 4B ^ 2 + 2) = 0 has real roots

Problems related to quadratic equation of one variable When a=_____ ,b=_______ The equation x ^ 2 + 2 (1 + a) x + (3a ^ 2 + 4AB + 4B ^ 2 + 2) = 0 has real roots

4(1+a)^2-4(3a^2+4ab+4b^2+2)>=0
4a^2+8a+4-12a^2-16ab-16b^2-8>=0
2a^2-2a+4ab+4b^2+1
Let's do it by ourselves...
It's hard to see mathematics.
According to the discriminant formula, in order to make the equation have real roots, we must make [2 (1 + a)] ^ 2-4 (3a ^ 2 + 4AB + 4B ^ 2 + 2) > = 0, and simplify to 2A ^ 2 + 4AB + 4B ^ 2-2a + 1
What is 4Y square + 8y + 3?
Cross multiplication:
21
23
Namely:
（2y+1）（2y+3）
The square of (2Y + 2) - 1
（2y+1 ）X（2y+3）
2x-3/x^2-1 - 3x^2-3/2x-3 =2
2x-3 / x ^ 2-1 minus 3x ^ 2-3 / 2x-3
Let a = (2x-3) / (x ^ 2-1), then (3x ^ 2-3) / (2x-3) = 3 / A, so A-3 / A-2 = 0A ^ 2-2a-3 = 0 (a + 1) (A-3) = 0A = - 1, a = 3 (2x-3) / (x ^ 2-1) = - 12x-3 = - x ^ 2 + 1x ^ 2 + x-4 = 0x = (- 1 ± √ 17) / 2 (2x-3) / (x ^ 2-1) = 32x-3 = 3x ^ 2-33x ^ 2-2x = x (3x-2) = 0x = 0, x = 2 / 3, so x = (- 1 - √ 17
Follow up questions
The distance between a and one or two people is 42km. If two people walk opposite each other at the same time, they will meet two hours later. If they have set out in the opposite direction, a will catch up immediately, and it will take 14 hours to catch up, so as to find the speed of a and a
Solve by equation
According to the question, the sum of the speed of a and B is 42 / 2 = 21 km / h
The speed difference between a and B is 42 / 14 = 3 km / h
Therefore, the speed of a is 1 / 2 (21 + 3) = 12 km / h
B speed is 1 / 2 (21-3) = 9 km / h
Compare the size of - 3x * 2 + 2x + 1 and - 3x * 2 + 2x-3x, please
-3x * 2 + 2x + 1 - (- 3x * 2 + 2x-3x) = 1 + 3x, so when x > - 1 / 3, - 3x * 2 + 2x + 1 > - 3x * 2 + 2x-3x when x = - 1 / 3, - 3x * 2 + 2x + 1 = - 3x * 2 + 2x-3x when x = - 1 / 3
The problem of "catching up" in mathematics application
Let's talk about the topic first: A and B start from the wharf at the same time in the same direction. B is in the front, 24 kilometers per hour. A is in the back, 28 kilometers per hour. Four hours later, a catches up with B, and how many kilometers are there between the two docks
The process is (28-24) * 4 = 16 km
My question is, isn't the above process seeking to catch up with the distance? Why is it related to the distance between the two docks? The title doesn't say that when we meet, B "just" arrives at the other dock
No score, but sincerely looking forward to your answer!
A_______ B____
Want to understand, as shown above
In the same direction, there is a saying that a overtakes B, which means that a and B were at two docks at the beginning
It is known that a = 2x ^ 2-3x-1, B = x ^ 2-3x-2 is to compare the size of a and B
A>B
A-B=2x^2-3x-1-(x^2-3x-2)=-x^2+1>0
So a > b
Find the size, do the difference method with more
A-B=x^2+1
x^2≥0 1>0
∴A-B>0
A>B
A minus B is always greater than 1, a is greater than 1
One variable equation of one degree
Party A and Party B are traveling in the same direction at a distance of 21 kilometers. Party A is behind and Party B is in front
It is known that a travels 8 kilometers per hour and B 5 kilometers per hour. How many hours does a start to catch up with B?
(please explain in detail, write the solution...)
As long as there is a process to solve this equation,
After 1.5 hours, the distance between a and B is 21-8 * 1.5 = 9
9 / (8-5) = 3 hours
3 + 1.5 = 4.5 hours
A: 4.5 hours
2x-1) ^ 2 (3x + 2) + (2x-1) (3x + 2) ^ 2-x (1-2x) · (3x + 2), where x = 1
X=1
(2x-1)^2(3x+2)+(2x-1)(3x+2)^2-x*(1-2x)*(3x+2) = 35
A primary school problem
The advance team of a PLA unit starts from the camp and advances to a certain place at the speed of 6 kilometers per hour. Six hours later, when there is an emergency in the army, the camera correspondent rides a motorcycle to contact at the speed of 78 kilometers per hour. How many hours later can the correspondent catch up with the advance team?
6*6=36
Let's catch up in an hour
36+6a=78a
a=0.5