On the condition that the equation x2-3x + M = 0 of X has two positive real roots

On the condition that the equation x2-3x + M = 0 of X has two positive real roots

x1>0,x2>0
Then x1x2 = m > 0
And the discriminant △ = 9-4m ≥ 0
So 0
B2-4ac is greater than or equal to 0
3x3-4x1xm greater than or equal to 0
Less than or equal to 9 / 4
If we solve the inequality system, we can find the range of M.
Of course, if you go to high school, there are other ways to solve this problem.
1. A class organizes extra-curricular activities to hold a badminton match, and plans to spend all 90 yuan on buying 2 badminton pairs
Rackets and more than 4 badminton. In the store, each racket is 25 yuan, and each ball is 2 yuan. In order to promote sales, the store has made a policy
There are two preferential schemes:
Scheme A: 10% discount on rackets and balls purchased;
Plan B: buy a racket and give 2 balls away
Customers can choose one of the options, but the two options cannot be used at the same time?
Equation of one degree with one variable!
Suppose you want to buy x balls
A 25 * 2 * 90% + 2 * 90% x = 90 45 + 1.8x = 90 1.8x = 45 x = 25
Type B: 25 * 2 + 2x = 90 2x = 40 x = 20 + 2 = 22
Answer to use a kind of plan to buy cost-effective
Set the ball x pair,
1. ：1.8x+22.5*2
2.： 2（x-2*2）+25*2
When 16
When 1 = 2, 1.8x + 45 = 2x + 42, so x = 6
Therefore, there are three situations to be answered separately.
When the ball bought (4
Factorization exercises
If f (x) = 6x ^ 3 + 11x ^ 2 + x-4 has a factor of X + 1, then what is the other factor of F (x)?
6x^2+5X-4
f(x)=(x+1)(6x^2+5x-4)
F (x) = 6x ^ 3 + 11x ^ 2 + x-4 = (x + 1) (6x ^ 2 + 5x-4), that is, another factor of F (x) is 6x ^ 2 + 5x-4
f(x)=6x^3+11x^2+x-4
=6x^3+11x^2+5x-4(x+1)
=x(6x^2+11x+5)-4(x+1)
=x(6x+5)(x+1)-4(x+1)
=(6x^2+5x+-4)(x+1)
=(2x+4)(3x-1)(x+1)
Another factor is 6x ^ 2 + 5x + - 4 or (2x + 4) (3x-1)
If the equation (x + 1 / xsquare-x) - (1 / 3x) = (x + K / 3x-3) with increasing roots, find K and X
(x+1/x²-x)-(1/3x)=（x+k/3x-3）
[3(x+1)-(x-1)]/3x(x-1)=x(x+k)/3x(x-1)
2x+4=x(x+k)
If the equation has an increasing root, then x = 0 or x = 1
When x = 0, K has no solution, so x ≠ 0
When x = 1, k = 5
To sum up, k = 5 x = 1
Two vehicles a and B are running towards each other from a and B at the same time. The meeting point of the two vehicles is 8km away from the midpoint of a and B. It is known that the speed of vehicle a is 1.2 times that of vehicle B, and the distance between a and B is calculated. (solution of equation)
In addition, question 9 on page 70 of the first volume of the first grade of junior high school
You can only answer the first question if you don't have one
Let the speed of car a be x, then car B be 1.2x, and let the time be t, then (1.2x-x) t = 8 * 2
Then x = 80 / T, the distance is s = (1.2x + x) * t = 176km
The factorization is as follows:
(1) xy^2+3xy-10x-y^2+4y-4
(2) (x+2)(x+3)(x-4)(x-5)-44
(1) The original formula = x (y ^ 2 + 3y-10) - (y ^ 2-4y + 4) = x (Y-2) (y + 5) - (Y-2) ^ 2 = (Y-2) (XY + 5x-y + 2) (2) we can use (x + 2) (x-4) and (x + 3) (X-5) to get (x ^ 2-2x-8) and (x ^ 2-2x-15), because the product of the two terms has the same term x ^ 2-2x, we need to find a way to eliminate the following
(1) xy^2+3xy-10x-y^2+4y-4
=x(y^2+3y-10)-(y^2-4y+4)
=x(y+5)(y-2)-(y-2)^2
=(y-2)(xy+5x-y+2）
(2) (x+2)(x+3)(x-4)(x-5)-44
=(x+2)(x-4)(x+3)(x-5)-44
=(x ^ 2-2x-8)... Expand
(1) xy^2+3xy-10x-y^2+4y-4
=x(y^2+3y-10)-(y^2-4y+4)
=x(y+5)(y-2)-(y-2)^2
=(y-2)(xy+5x-y+2）
(2) (x+2)(x+3)(x-4)(x-5)-44
=(x+2)(x-4)(x+3)(x-5)-44
=(x^2-2x-8)(x^2-2x-15)-44
=(x^2-2x-23/2+7/2)(x^2-2x-23/2-7/2)-44
=(x^2-2x-23/2)^2-.49/4-44
=(x^2-2x-23/2)^2-.225/4
=(x^2-2x-23/2+15/2)(x^2-2x-23/2-15/2)
=(x ^ 2-2x-4) (x ^ 2-2x-19) fold up
（1）＝（y-2)(xy+5x-y+2)
(2)=(x^2-2x-4)(x^2-2x-19)
If the equation (x + 1) / (X & sup2; - x) - 1 / 3x = K / (3x-3) about X has an increasing root, try to find the increasing root of the equation and the value of K
The denominator of CM is 3x (x-1), and both sides of the equation are multiplied by 3x (x-1) at the same time
3（x+1)-（x-1)=kx
Root increment can only be 0 or 1
Substituting x = 0 into the above formula, we get: 3 + 1 = 0, which is inconsistent
Substituting x = 1 into the above formula, we get: 6-1 = k, we get: k = 5
So the increasing root is 1, k = 5
Solving several practical problems with equations
It took 55 minutes for someone to go down the mountain at a speed of 12 kilometers per hour and pass through the flat road B at a speed of 9 kilometers per hour. When he came back, he passed the flat road at a speed of 8 kilometers per hour and went up the mountain at a speed of 4 kilometers per hour. It took 1.5 hours to get the distance between a and B
When m is a value, the solution of the equation 5m + 12x = 1.5 + X about X is 2.0 times larger than that of the equation x (M + 1) = m (1 + x)
Suppose the slope is x and the level is y, then x / 12 + Y / 9 = 55 / 60x / 4 + Y / 8 = 1.5, the solution is x = 3Y = 63 + 6 = 9 (km), so the distance between a and B is 9 km. 5m + 12x = 1.5 + x11x = 1.5-5mx = (1.5-5m) / 11x (M + 1) = m (1 + x) MX + x = m + mxx = m, because the solution of the equation 5m + 12x = 1.5 + X about X is better than the solution of the equation
It's exactly right upstairs
On the top floor
By solving the equation, x = 3, y = 6, x + y = 9, the distance is 9 km
The answer of factoring exercises (junior high school questions)
① X & # 178; - X-30 ② x-29x-30 ③ X & # 178; + X-30 ④ X & # 178; + 7x-30 ⑤ X & # 178; - 7x-30 ⑥ X & # 178; - 13x-30 ⑦ X & # 178; + 13x-30 Ⅷ X & # 178; + 7x-30 Ⅸ X & # 178; - X-30 question 10: X & # 178; - X-12
① (x + 5) (X-6) 2 (x + 1) (X-30) 3 (X-5) (x + 6) 4 (x-3) (x + 10) 5 (x + 3) (X-10) 6 (x + 2) (X-15) 7 (X-2) (x + 15) 8 same as 4 9 same as 1 (x + 3) (x-4)
The equation (x + 1) / (x ^ 2-x) - 1 / 3x = (x + k) / (3x-3) of X has increasing roots. Find out the values of increasing roots and K respectively
X (x-1) = 0, x = 0, 3 (x-1) = 0. Why?
(x + 1) / (x ^ 2-x) - 1 / 3x = (x + k) / (3x-3) multiply both sides by 3x (x-1) 3 (x + 1) - (x-1) = x (x + k) x ^ 2 + KX = 2x + 4. The increasing root means that the denominator is 0, so x ^ 2-x = 0,3x = 0,3x-3 = 0x (x-1) = 0, x = 0,3 (x-1) = 0. So the increasing root may be x = 0, x = 1. Substituting x = 0, x = 1 into x ^ 2 + KX = 2x + 40 + 0 = 2 * 0 + 4 does not hold, so x