Quadratic equation of one variable 1. It is known that equation 2-root 5 is one root of equation xsquare-4x + C = 0, then what is the other root? 2. Solve the equation. Add 1 to each root of equation xsquare-4x-3 = 0; Then the new equation is? 3. If the equation AX ^ 2 + X-1 = 0 about X has real roots, then the range of a is 4. It is known that x = 1 is a solution of the quadratic equation AX ^ 2 + bx-40 = 0 of one variable, and a is not equal to B. find the value of a ^ 2-B ^ 2 divided by 2a-2b. 5. It is known that the equation x ^ 2-2 (M + 1) XM ^ 2 = 0 about X (1) When m takes what value, the equation has no real root (2) take a suitable non-zero integer for M. make the equation have two real roots, and find the square of their differences. 6. Given that x1.x2 is the relationship between two utilization coefficients and roots of 2x ^ 2 + 4x-3 = 0, find the value of x2 / X1 + X1 / x2

Let me introduce you some theorems, a quadratic equation: ax ^ 2 + BX + C = 0 (a ≠ 0)
Theorem 1
If there are two real roots, it must satisfy △ = B ^ 2-4ac ≥ 0. When we take the equal sign, the two real roots are equal (this does not mean that there is only one real root, but the two real roots are equal). When △ = B ^ 2-4ac < 0, we say that it has no solution in the real range. Of course, what you are learning now should be in the default real range, In fact, it has a solution in the range of complex number, which is not discussed here
Theorem 2
When the equation has two real roots (no matter whether the two roots are equal or not, this is the premise. If there is no real root, the theorem can not be used, because the complex number is not discussed at present, although it still holds in the range of complex number). Let the two roots be X1 and X2 respectively, then the equation X1 + x2 = - B / A, x1x2 = C / A is satisfied
Next, it's easy to do these questions. You can try it yourself. If you still can't do it, I'll see the answers below
Writing is a good habit for middle school students
∵ X1 = 2 - √ 5, and X1 + x2 = 4
∴ x2 = 2+√5.
∵ x1 + x2 = 4 ,x1 x2 = -3;
∴ (x1 + 1) + (x2 + 1) = 6 ,(x1 + 1)(x2 + 1) = x1 x2 + x1 + x2 +1 = 2
The new equation is x ^ 2 - 6x + 2 = 0
∵ a may be equal to 0, so first discuss whether a = 0 satisfies the meaning of the question
(at this time, the equation is not a quadratic equation, but a linear equation, so it must be discussed separately.)
It is easy to know that a = 0 satisfies the meaning of the question;
When a ≠ 0:
The expression on the right side of a ∈ [- 1 / 4, + ∞) is equivalent to a ≥ - 1 / 4
∵ take x = 1 into the equation to get a + B - 40 = 0, that is, a + B = 40;
Qi
According to the square difference formula, (a ^ 2 - B ^ 2) / 2 (a - b) = (a + b) (a - b) / 2 (a - b)
= (a + b)/2 = 20 .
(1) (personally, I think you missed the plus sign.)
∵ △ = 4（m + 1)^2 - 4m^2 = 4(2m + 1)
When the equation has no real root, the solution is m < - 1 / 2
(2)
∵ （x1 - x2)^2 = (x1 + x2)^2 - 4x1x2 = 8m + 4;
When m = 1, the result is 12
x2/x1 + x1/x2 = (x1^2 + x2^2)/x1x2 = [(x1 + x2)^2 - 2x1x2]/x1x2
= (x1 + x2)^2/x1x2 - 2 = -14/3.
When the formula is not proficient enough, it's better not to wait as I do, because when reading the paper, it's divided equally by step. If the result is wrong, it will be deducted completely. If it's disassembled, even if the result is wrong, there will still be scores in the process, so generally don't wait, I just want to make it convenient
For the quadratic equation AX & # 178; + BX + C = 0, there is the following formula
Two sum = - B / A, two product = C / A,
The discriminant of root = B & # 178; - 4ac, > 0, there are 2 unequal roots, = 0, there are 2 equal roots (i.e. a root), 0, there are 2 unequal roots, = 0, there are 2 equal roots (i.e. a root), = 0, a
Eight people take two cars with the same speed to the railway station at the same time. Each car takes four people (excluding the driver). One of the cars breaks down 10 kilometers away from the railway station. At this time, there is still 28 minutes to stop checking. At this time, the only available traffic tool is another car. It is known that this car, including the driver, can only take five people, and this car can not stop checking tickets The average speed of the car is 60 km / h, while the average speed of people walking is 5 km / h. can these eight people get to the railway station before the check-in stops?
When the car breaks down, four people in this car get off and walk, another car will take four people in its car to the station, and then come back to pick up the four people walking, and then send them to the railway station; suppose that the distance of four people in this car get off and walk is x, according to the meaning of the question: X5 = 10 + 10 − X60, the solution is x = 2013, then the time of four people walking to the station is: 2013 △ 5 + (10-2013) △ 60 = 3578 (small) A: these eight people can get to the railway station before the check-in stops
If a root of the quadratic equation (m-1) x + 5x + m-3m + 2 is 0, then M =? 2. The quadratic equation k * x-2x + 1 has a real root, then what is the value range of K? 3
The third mistake is a ^ 2-B ^ 2
My friend, you haven't finished the first problem equation, and finally = 0. The solution is a root of 0. If you take this root into this equation, you can get the square of M - 3M + 2 = 0, then you can get m = 1, M = 2
2. You are not an equation yet. The solution is as follows: the square of kx-2x + 1 = 0 has roots. Then, when k = 0, x = 1 / 2 holds. When k is not equal to 0, if there are roots, then the square of b-4ac is greater than or equal to 0 and K is less than or equal to 1. So the final answer is k is less than or equal to 1.3. Did you copy the third question wrong
1. When x = 0, the left side of the equation is 0, so m ^ 2-3m + 2 = 0, and the solution is m = 2 or M = 1. Because the equation is a quadratic equation with one variable, M-1 cannot be equal to 0, but M = 2
2。 If there is a real root, then B ^ 2-4ac ≥ 0, that is, 2 ^ 2-4 * 1 * k ≥ 0, K ≤ 1 and K ≠ 0 (quadratic equation of one variable)
3。 Are you sure it's not a ^ 2-B ^ 2?
1、
Bring x = 0 in
M^2-3M+2=0
M = 2 or 1
2. The range of K is (- ∞, + ∞)
3. Bring x = 1 into
a+b=40
(a^2+b^2)/(2a-2b)
=
Bring x = 0 in
M^2-3M+2=0
M = 2 or 1
The 36 cm long wire is cut into two equal sections. One section is used to form a rectangle, and the other section is used to form an isosceles triangle with one side of 5 cm. If the area of the rectangle is equal to the area of the triangle, calculate the side length of the rectangle
I've been raised to 20, no one?
The perimeter of an isosceles triangle and a rectangle is 18. If one side of the rectangle is a, the other side is 9-A, and the area of the rectangle is a * (9-A)
In the first case, if the two waists of an isosceles triangle are 5, then the third side is 8. If you make the vertical line of the third side, then the height is 3. So the area of the triangle is 12
So a * (9-A) = 12
So a * A-9 * a + 12 = 0
The two roots are (9-radical 33) / 2 and (9 + radical 33) / 2, which are the length of two sides
The second case: if the third side of an isosceles triangle is 5, then the two waists are 6.5. Do the same as above, but this is more complicated. Do it yourself
The length of the rectangle is widened to 9, and the area is equal to that of the isosceles triangle to 12. Therefore, it is to solve a quadratic equation of two variables, y ^ 2-9y + 12 = 0, and calculate the side length of the rectangle
Side length is a, width is 18-a, triangle isosceles, waist length is 5, low is 18-5 * 2 = 8, triangle area is
S = 0.5 * 8 * 3 = 12 is equal to the rectangular area, that is, a (18-a) = 12
I don't know how to ask
1.3^2004-4*3^2003+10*3^2002=______
2.3^n+2-4*3^n+1+10*3^n=______
3. The length of a rectangle is am and the width is BM (a > b). If the length is increased by 2M and the width is decreased by 3M, then the area of the rectangle is increased or decreased? How many square meters are increased or decreased?
1. Original formula = 3 * 3 ^ 2003-4 * 3 ^ 2003 + 10 * 3 ^ 2002
=-3^2003+10*3^2002
=-3*3^2002+10*3^2002
=7*3^2002
2. The original formula = 3 * 3 ^ (n + 1) - 4 * 3 ^ (n + 1) + 10 * 3 ^ n
=-3^(n+1)+10*3^n
=-3*3^n+10*3^n
=7*3^n
3.s=ab-(a+2)(b-3)=ab-(ab-3a+2b-6)=3a-2b+6
Because a > b,
So s > 0, reduce 3a-2b + 6 square meters
1、7*3^2002
2、7*3^n
3. Less. 3a-2b + 6 is reduced
1.7*3^2002
2.7*3^n
3. Decrease, 3a-2b + 6
X & # 178; - 2 √ 3x = - 3 solve the equation
x²-2√3x+3=0
(x-√3)²=0
x-√3=0
x=√3
In the third year of junior high school mathematics, we use the formula method to solve the following quadratic equation of one variable
6x²-13x-5=0
x（x+8）=16
-½-3x+6=0
6x²-13x-5=0
x=[13±√(169+120)]/12
x=(13±17)/12
x=5/2,x=-1/3
x（x+8）=16
x^2+8x-16=0
x=[-8±√(64+64)]/2
x=-4±4√2
-½x^2-3x+6=0
x=[3±√(9+12)]/(-1)
x=-3±√21
If a square-3a + 1 = 0, then a square + 1 / a square =? (A-1 / a) & sup2; =?
A square - 3A + 1 = 0 (divide a by a)
a+1/a=3
A square + 1 / a square
=A square + 1 / a square + 2-2
=(a + 1 / a) square-2
=9-2
=7
(A-1 / a) square
=A square - 2 + 1 / a square (above a square + 1 / a square = 7)
=7-2
=5
3x & # 178; - (x + 2) &# 178; + 2x = 0
3x²-(x+2)²+2x=0
3x²-x²-4x-4+2x=0
2x²-2x-4=0
x²-x-2=0
(x-2)(x+1)=0
x1=2 x2=-1
If a project is completed within 6 days by team a and team B, the manufacturer shall pay 8700 yuan to team a and team B; if team B and team C complete within 10 days, the manufacturer shall pay 9500 yuan to team B and team C; if team a and team C complete 23 projects within 5 days, the manufacturer shall pay 5500 yuan to team a and team C. (1) how much does it take for each team to complete all projects separately? (2) If this project is required to be completed within 15 days, which project can be completed at least? Please give reasons
(1) Let team a do it alone in X days, team B do it alone in y days, and team C do it alone in Z days. Then, solve the equation system in LX + 1y = 161y + 1z = 1101x + 1z = 23 × 15, and get ∧ x = 10Y = 15z = 30 (2) team C do it in 30 days first

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