How much is it to change the equation 1 / 2 (x-1) ^ 2 + 3x = 5 / 2 into the general form of quadratic equation of one variable? I want the general form of quadratic equation of one variable, don't work it out! )

How much is it to change the equation 1 / 2 (x-1) ^ 2 + 3x = 5 / 2 into the general form of quadratic equation of one variable? I want the general form of quadratic equation of one variable, don't work it out! )

1/2(x-1)^2+3x=5/2
(x-1)^2+6x-5=0
x^2-2x+1+6x-5=0
x^2+4x-4=0
x^2+4x-4=0
Change the equation 3x (x-1) = x (x + 2) + 8 into general form, and write out the coefficients of quadratic term, coefficients of primary term and several constant terms
General form 2x ^ 2-5x-8 = 0, quadratic coefficient 2, primary coefficient - 5, constant - 8, I hope my answer can help you
Just unfold
3x²-3x=x²+2x+8
2x²-5x-8=0
Quadratic coefficient: 2
Coefficient of first term: - 5
Constant term; - 8
3x（x-1）=x（x+2）+8
3x²-3x=x²+2x+8
2x²-5x-8=0
The quadratic coefficient is 2
The coefficient of the first term is - 5
The constant term is - 8
The application of 1-variable quadratic equation in the third grade mathematics
Someone will hold a cup of 63 liters of pure caustic soda solution, pour out a small cup, fill up with water, stir evenly, and pour out the same small cup of solution, the container also has 28 liters of pure caustic soda solution, ask him the volume of the small cup he used. To solve the problem in detail, not only the formula and the number
Let the volume of the small cup he uses be x liters
After the first pour out, there are still 63-x liters of pure liquid, and the concentration of the solution is 0
（63-x)/63
Pour out the pure liquid x (63-x) / 63 for the second time
∴63-x-x（63-x)/63=28
X1 = 21, X2 = 105
A: the volume of the small cup he used is 21 liters
Just looking at it, you can see that this cup is not small at all
If the cup is easy to be x-liter, there will be 63-x after each pour
Now with water, the concentration is (63-x) / 63
Once more, the purity of the reverse is (63-x) / 63 * X
So there are
63－X-（63－X)/63*X=28
35-2X+XX/63=0
X^2-126X+2205=0
X1 = 115, rounded off x2 = 31
The volume is 31 liters... Spread out
Just looking at it, you can see that this cup is not small at all
If the cup is easy to be x-liter, there will be 63-x after each pour
Now with water, the concentration is (63-x) / 63
Once more, the purity of the reverse is (63-x) / 63 * X
So there are
63－X-（63－X)/63*X=28
35-2X+XX/63=0
X^2-126X+2205=0
X1 = 115, rounded off x2 = 31
The volume is 31 liters
Find the formula of meeting problem and chasing problem
Basic concept: the travel problem is to study the motion of objects, it studies the relationship among the speed, time and travel of objects. Basic formula: distance = speed × time; distance △ time = speed; distance △ speed = time. Key problem: determine the position in the process of travel encounter problem: speed and × encounter time
The line 3x-4y-5 = 0 and the circle C (X-2) ^ 2 + (Y-1) ^ 2 = 25 intersect at two points a and B, and calculate the area of △ ABC
The center of circle C (X-2) ^ 2 + (Y-1) ^ 2 = 25 is C (2,1), the radius is 5, and the distance from the center to the straight line is d = | 3 × 2-4 × 1-5 | / √ (3 & # 178; + 4 & # 178;) = 3 / 5 half chord length AB / 2 = √ [5 & # 178; - (3 / 5) & # 178;] = 2 √ 154 / 5ab = 4 √ 154 / 5S = abd / 2 = (4 √ 154 / 5 × 3 / 5) / 2 = 6 √ 154 / 25
The formula of encounter problem and pursuit problem
Set the distance as s, the speed as V1 and the time as t
Encounter time = s / (V1 + V2)
Encounter distance = (V1 + V2) * t
Pursuit time = s / (V1-V2)
Set the distance as s and the speed as V1 and V2
Encounter time = s / (V1 + V2)
Pursuit time = s / (V1-V2)
The straight line 3x-4y-5 = 0 intersects the circle C (X-2) + (Y-1) = 25 at two points a ` B to find the area of triangle ABC
What is point C? Is it a triangle ABO?
If so, we can use the coordinate method
Firstly, the distance from O to line, that is, the height of triangle, is calculated by using the formula of distance from point to line
Then the linear equation is brought into the circular equation to get the coordinates of a and B. then the length of AB, the base of triangle, is obtained
Finally, the area of the triangle can be calculated
It's very simple. First draw a picture. The radius is 5. The distance from the center of the circle to the straight line is high = [6-4-5] / 5 = 3 / 5
Half of Xuan's length is obtained by Pythagorean theorem, and then * 2 is the bottom of the triangle.
Distance problem: S = VT encounter problem: pursuit problem: engineering problem: profit problem:
Encounter problem encounter distance = velocity sum × encounter time encounter time = encounter distance △ velocity sum = encounter distance △ encounter time catch up problem catch up distance = velocity difference × catch up time catch up time = catch up distance △ velocity difference velocity difference = catch up distance △ catch up time
What is the specific topic? Follow up: application problems generally have distance problems: S = VT encounter problems: follow up problems: engineering problems: profit problems: no more
It is known that the line 3x + 4y-12 = 0 intersects the x-axis and y-axis at two points a and B, and the point C moves on the circle (X-5) 2 + (y-6) 2 = 9, then the difference between the maximum and minimum value of △ ABC area is 0______ .
Let a line parallel to a known line and tangent to circle (X-5) 2 + (y-6) 2 = 9 be made, and the tangent points are P1 and P2 respectively. As shown in the figure, when moving point C moves on circle (X-5) 2 + (y-6) 2 = 9, if C coincides with point P1, △ ABC area reaches the minimum; when C coincides with point P2, △ ABC area reaches the maximum ∵ line 3x + 4y-12 = 0 intersects with X axis and Y axis at a (4,0) and B (0,3) The difference between the maximum and minimum values of | ab | = 42 + 32 = 5 | ABC area is s = s △ abp2-s △ ABP1 = 12 | ab | (d2-d1) = 52 (d2-d1), where D2 and D1 are the distance between point P2 and line AB respectively ∵ P1 and P2 are two parallel tangents of circle (X-5) 2 + (y-6) 2 = 9 | the difference between the distance between point P2 and point P1 and line AB is equal to the diameter of circle, that is, d2-d1 = 6, so the maximum and minimum values of | ABC area are obtained The difference of small value is 52 (d2-d1) = 52 × 6 = 15, so the answer is: 15
Is there a formula for chasing and meeting problems?
The basic quantity relations of travel problems are as follows:
Speed × time = distance
Distance △ speed = time
Distance △ time = speed
1. Encounter problem:
Sum of speed × time of encounter = distance between two places
Distance between two places △ sum of speed = time between two places
Distance between two places △ meeting time = sum of speed
2. Follow up questions:
Pursuit distance △ speed difference = pursuit time
Speed difference × pursuit time = pursuit distance
Pursuit distance △ pursuit time = speed difference
Fast slow = speed difference
Pursuit problem
Speed difference * time = distance between the two
Distance / speed difference = time
Distance / time = speed difference
Encounter problem
Speed and * time = distance
Distance / speed and = time
Distance / time = speed and speed
Encounter problem
Meeting distance = speed of two cars and meeting time
Meeting time = meeting distance △ speed of two cars and
Speed sum of two cars = distance of meeting △ time of meeting
Follow up questions
Overtaking distance = speed difference between two vehicles × overtaking time
Overtaking time = overtaking distance △ speed difference between two vehicles
Speed difference between two vehicles = overtaking distance △ overtaking time