The equation (X-2) (3x-5) = 1 is reduced to the general form () where a =, B =, C=

The equation (X-2) (3x-5) = 1 is reduced to the general form () where a =, B =, C=

x(3x-5)-2(3x-5)=1
3x²-5x-6x+10=1
3x²-11x+9=0
a=3,b=-11,c=9
3x square - 11x + 9 = 0
A=3
b=-11
The equation (3x-1) 2 - (x-3) 2 = x + 3 is reduced to a general form
(3x-1)2-(x-3)2=x+3
9x²-6x+1-x²+6x-9-x-3=0
8x²-x-11=0
If you don't understand this question, you can ask. If you are satisfied, please click "praise"
Examples of the problem of "first" pursuing "second" meeting in Mathematics in grade one
For example: ① A and B start at the same point, a run 3 meters first, B catch up, B catch up with a in a few hours? ② B and B walk opposite each other at the same point, they meet in a few hours? (A's speed is 20 km / h, B is 25 km / h) ① solution B x hours catch up with A. 3 + 20x = 25X ② solution
Chasing first means running in circles.
Suppose a is in the front and B is in the back. A is 5 meters faster than B every second. A circle of runway is 400 meters. Two people start at the same time. How many seconds does it take a to catch up with B?
If the front one wants to catch up with the back one, he has to run one more lap than the back one, so 400 △ 5 = 80 seconds
A is in the front, B is in the back, two people start at the same time, catch up with B at the end of the third lap, each lap of the runway is 400 meters, how much faster is a than B per second?
Three laps 1200 meters, a runs three laps, B runs two laps (catch up with B), so (1200-800) △ 800 = 0.5 meters
All real roots of equation x2 + 3x-2 = 0
x^2+3x+(3/2)^2-(3/2)^2--2=0
(x-3/2)=13/4
x=-√13/2+3/2 x=+√13/2+3/2
quadratic formula
X = - B + √ B ^ 2-4ac / 2A or x = - B - √ B ^ 2-4ac / 2A
-3 + √ 13 / 2 or - 3 - √ 13 / 2
Find the formula of pursuit problem and encounter problem!
RT
The pursuit problem and the encounter problem are equal in distance
Pursuit problem: distance = speed difference × pursuit time
Encounter problem: distance = speed and X encounter time
W both the pursuit problem and the encounter problem are equal in distance
Pursuit problem: distance = speed difference × pursuit time
Encounter problem: distance = speed and X encounter time
On the equation of X, if one root of the square of X + 3x + k = 0 is - 1, then k = (), and the other root is ()
From the meaning of the title, we can see that X & # 178; + 3x + k = 0
x²+2×（2/3x)+K=0
x²+2×（2/3x)+4/9-4/9+K=0
(x+2/3)²-4/9+K=0
(x+2/3)²=4/9-K
From the meaning of the title, we can see that a root is - 1, and take it with the above formula (- 1 + 2 / 3) &# = 4 / 9-k, that is, k = 2
Substituting k = 2 into (x + 2 / 3) ² = 4 / 9-k
(x + 2 / 3) & #178; = 4 / 1, then X1 = (- 1) x2 = (- 2)
On the equation of X, if one root of the square of X + 3x + k = 0 is - 1, then k = (2), and the other root is (- 2)
The other root is a
Then a + (- 1) = - 3
a*(-1)=k
∴a=-2
K=2
X ^ 2 + 3x + k = 0, X1 = - 1, k = 2, x1 × x2 = C / A, X2 = - 2.
What are the formulas of the two problems?
Follow up questions:
Pursuit distance = speed difference × pursuit time
Time to catch up = distance to catch up △ speed difference
Encounter problem:
Encounter distance = speed and X encounter time
Encounter time = encounter distance △ speed and
What is the maximum distance between the point on circle (x-1) ^ 2 + (Y-2) ^ 2 = 9 and the line 3x + 4y-15 = 0
According to the standard equation of a circle, the center of the circle is (1,2), and the radius is 3. The center of the circle does not satisfy the linear equation, so the center of the circle is not on the line. There are two intersections between the line and the circle. Therefore, the maximum distance from the line to a point on the circle should be the distance from the center of the circle to the line plus a radius
The distance from the center of the circle to the straight line is | 3 * 1 + 4 * 2-15 | / 5 = 3 / 5
The maximum is this number plus radius equal to 3 + 3 / 5
The formula of pursuit and encounter
The formula of pursuit problem is: speed difference times pursuit time = pursuit distance
The formula of encounter problem is: speed sum times time = distance sum
The maximum distance between the point on the circle x ^ 2 + y ^ 2 = 1 and the line 3x + 4y-25 = 0 is________
Can't this problem be wrong=
The maximum distance from point to line is 6