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Bivariate quadratic system of application problems, listed bivariate linear equations and solutions 1. There are three rooms and two rooms in a hotel. The three rooms are 25 yuan per person per day, and the two rooms are 35 yuan per person per day. It costs 1510 yuan for a tour group of 50 people to stay in the hotel. How many rooms do you rent for each of the two rooms? 2. A stadium runway 400 meters, a, B reverse line, meet once every 30 seconds, the same direction, every 80 seconds B catch up with a, a, B speed respectively?
Let 1 rent 3 people x, 2 people y, 3x + 2Y = 50, 1 3 × 25X + 2 × 35y = 1510, 2 1 × 35: 105x + 70Y = 1750, 3 3 - 2 get: 30x = 240, x = 8, 1 get: y = 13, so 3 people 8, 2 people 13
Exercises of quadratic equation of one variable
There are four equations as follows
(1)x^ - 2x- 2=0(2)2x^ +3x -1=0(3)2x^ - 4x +1=0 (4)x^+6x + 3=0
The coefficients of the first term of three equations have the same characteristics
Please use algebraic expression to express this characteristic and deduce the root formula of quadratic equation with one variable
There are four equations as follows
(1)x^ - 2x- 2=0(2)2x^ +3x -1=0(3)2x^ - 4x +1=0 (4)x^+6x + 3=0
The coefficients of the first term of three equations have the same characteristics
Please use algebraic expression to express this characteristic and deduce the root formula of quadratic equation with one variable
Equation (1) X1 + x2 = 2
Equation (2) X1 + x2 = - 3 / 2
Equation (3) X1 + x2 = 2
Equation (4) X1 + x2 = - 6
Common ground: the sum of the two roots is an integer. (equation (1) (3) (4)), which can be matched in the form of complete square
Namely: (x + m) ^ 2 = n
The root is: x = - m, the root is n
Solving five exercises of quadratic equation of one variable
The simplest one, a formula
I found 20,
（1）(3x+1)^2=7
(3x + 1) ^ 2 = 7 ℅ (3x + 1) ^ 2 = 7 ℅ 3x + 1 = ± √ 7 (be careful not to lose the solution) ℅ x = (± √ 7-1) / 3
（2）9x^2-24x+16=11
The solution of the original equation is X1 = (√ 11 + 4) / 3 x2 = (- √ 11 + 4) / 3
(3) (x+3)(x-6)=-8
(x + 3) (X-6) = - 8 is reduced to x ^ 2-3x-10 = 0 (the left side of the equation is a quadratic trinomial, and the right side is zero) (X-5) (x + 2) = 0 (the left side of the equation is a factorization factor) ‖ X-5 = 0 or x + 2 = 0 (converted into two univariate linear equations) ‖ X1 = 5, X2 = - 2 are the solutions of the original equation
(4) 2x^2+3x=0
2X ^ 2 + 3x = 0, X (2x + 3) = 0 (decompose the left side of the equation with the method of raising the common factor); X = 0 or 2x + 3 = 0 (transform into two unary linear equations); X1 = 0, X2 = - 3 / 2 are the solutions of the original equation. Note: it is easy for some students to lose the solution of x = 0 when doing this kind of problem, so we should remember that unary quadratic equation has two solutions
(5) 6X ^ 2 + 5x-50 = 0
6x2 + 5x-50 = 0 (2x-5) (3x + 10) = 0 (pay special attention to the sign not to make mistakes when factoring by cross multiplication) ‖ 2x-5 = 0 or 3x + 10 = 0 ‖ X1 = 5 / 2, X2 = - 10 / 3 is the solution of the original equation
(6) X ^ 2-4x + 4 = 0
X ^ 2-4x + 4 = 0 (∵ 4 can be decomposed into 2.2, this problem can be solved by factorization method) (X-2) (X-2) = 0 ∵ X1 = 2, X2 = 2 is the solution of the original equation
(7)（x-2）^2=4（2x+3）^2
Solution. (X-2) ^ 2-4 (2x + 3) ^ 2 = 0. [X-2 + 2 (2x + 3)] [(x-2-2 (2x + 3) = 0
(5x+4)(-5x-8)=0.
x1=-4/5,x2=-8/5
（8）y^2+2√2y-4=0
Solution (y + √ 2) ^ 2-2-4 = 0
(y+ √2)^2=6.
y+√2=√6.
y=-√2±√6.
y1=-√2+√6；
y2=-√2-√6.
（9）（x+1）^2-3（x+1)+2=0
Solution (x + 1-1) (x + 1-2) = 0
x(x-1)=0.
x1=0,
x2=1.
(10) X ^ 2 + 2ax-3a ^ 2 = 0 (a is constant)
The solution (x + 3a) (x-a) = 0
x1=-3a,
x2=a.
(11)2x^2＋7x＝4．
The equation can be changed to 2x ^ 2 + 7x-4 = 0
∵a＝2,b＝7,c＝－4,b2－4ac＝72－4×2×(－4)＝81＞0,
∴x＝ ．∴x1＝ ,x2＝－4．
(12)x^2－1＝2 x
The equation can be changed to x ^ 2-2, X-1 = 0
∵a＝1,b＝－2 ,c＝－1,b2－4ac＝(－2 )2－4×1×(－1)＝16＞0．
∴x＝ ．∴x1＝ ＋2,x2＝ －2
（13） x^2 + 6x+5=0
The original equation can be reduced to (x + 5) (x + 1) = 0
x1=-5 x2=-1
(14) x ^2－4x+ 3=0
The original equation can be reduced to (x-3) (x-1) = 0
x1=3 x2=1
(15)7x^2 －4x－3 =0
The solution of the original equation can be reduced to (7x + 3) (x-1) = 0
x1=-3/7 x2=1
(16)x ^2－6x+9 =0
The solution of the original equation can be reduced to
(x-3)^2=0
x1=x2=3
(17)x²+8x+16=9
(x+4)²=9
X + 4 = 3 or x + 4 = - 3
x1=-1,x2=-7
(18)(x²-5)²=16
X & # 178; - 5 = 4 or X & # 178; - 5 = - 4
X & # 178; = 9 or X & # 178; = 1
x1=3,x2=-3,x3=1,x4=-1
(19)x(x+2)=x(3-x)+1
Solution X & # 178; + 2x = 3x-x & # 178; + 1
2x²-x-1=0
(2x+1)(x-1)=0
x1=-1/2 x=1
(20) 6x^2+x-2=0
The solution of the original equation can be reduced to (3x + 2) (2x-1) = 0
(x+2/3)(x-1/2)=0
x1=-2/3 x2=1/2
As shown in the figure, in the equilateral triangle ABC, O is a point in the shape, the angle AOB = 105 ° and the angle AOC = 125 °. Take OA, OB and OC as the internal angle of the triangle formed by the edges
Rotate the triangle ABO by 60 degrees counter clockwise with B as the center to o'bc, and the AB and CB sides coincide to connect OO '
The obvious triangle oo'b is an equilateral triangle
OO'=OB
O'C=OA
So the lengths of the three sides of the triangle OCO 'are OA, OB and OC respectively
The ratio of AOB to BOC is 6:5:4
So BOC = 5 × 24 = 120 degrees
Angle AOB = angle bo'c = 6 * 24 = 144 degrees
And the angle boo '= angle bo'o = 60 degrees
So the angle o'oc = 120-60 = 60 degrees
Angle oo'c = 144-60 = 84 degrees
Angle OCO '= 180-60-84 = 36 degrees
Ao, Bo and Co are the angles of each side of the triangle, which are 60 degrees, 36 degrees and 84 degrees
As shown in the figure below, take OA as the hypotenuse to make the isosceles right triangle OAB, and then take ob as the hypotenuse to make the isosceles right triangle OBC on the outside of △ OAB
When it comes to 8 isosceles right triangles, what is the area ratio of the triangle OAB to the last triangle? It's a process (I'm just a freshman, don't use other processes that I don't understand, OK, give me points)
If there are 8 isosceles right triangles, the area ratio of △ OAB and △ OHI is ()
A. 32\x09B. 64\x09C. 128\x09D. 256
Urgent need
To go to the eighth isosceles right triangle is to divide the angle equally and then divide the rest equally for eight consecutive times. For a simple example, how much is half of half? Half of half is equal to half divided twice, which is 1 / 4 1:4, and the original number is the square of 2
Similarly, the result is the square of 8, 8 * 8 = 64
For example, when k ＞ 0 and B ＞ 0, the image passes through the first, second and third quadrants. Why? Reason + 100
① First confirm the direction of the straight line: k > 0, that is to say, the larger x, the larger y, that is to say, the straight line is inclined upward!
② Then calculate the intercept between the line and the y-axis
y=kx+b,
Intercept of line with y axis: that is, x = 0, so y = KX + B = 0 + B = B, b > 0
The intercept is greater than zero, which means that the line intersects the positive half axis of the y-axis,
The positive half axis of the intersecting y-axis, plus it is inclined upward: draw through the coordinate axis, you can know that the image passes through 1, 2, 3 quadrants
The image of function y = K (x-k) (k < 0) does not pass through______ Quadrant
The function y = K (x-k) = kx-k2. ∵ K < 0, the image passes through the second and fourth quadrants; ∵ - K2 < 0, the intersection of the image and the Y axis is below the X axis, and the image also passes through the third quadrant; therefore, the image of the function y = K (x-k) (k < 0) does not pass through the first quadrant
How many quadrants does the image of a function y = (k ^ 2 + 1) x pass through!
K ^ 2 + 1 > 1 > 0, so y and X have the same sign
If the first-order function y = 2 (1-k) x + 2 / 1k-1 does not pass through the first quadrant, then the value range of K is?
Write process!
If the linear function y = 2 (1-k) x + 2 / 1k-1, the image does not pass through the first quadrant,
Slope 2 (1-k) 1,
What is 2 / 1k-1?
If the image of a linear function y = (K + 1) x + K-2 does not pass through the second quadrant, then the value range of K is______ ．
∵ if the image of the first-order function y = (K + 1) x + K-2 does not pass through the second quadrant, ∵ K + 1 > 0, K-2 < 0, the solution is - 1 < K ≤ 2, and the value range of ∵ K is - 1 < K ≤ 2
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- 11. 1. Construct a system of quadratic equations of two variables by image method, and make its corresponding linear function pass through (- 3, - 2), (- 1,6) two points 2. Given that the image intersection of the first-order function y = 2x-a and y = 3x + B and a point outside the origin on the x-axis, what is a / (a + b) equal to?
- 12. If f (x) = 4, what is the value of X
- 13. The known function f (x) = 2x-1 g (x) = {X & # 178;, X ≥ 0} (piecewise function) - 1, x < 0 Why do we need to find the norm of X again to find the analytic expression of G [f (x)] g（x）={x²，x≥0 } -1，x＜0
- 14. (2) given y = f (x) = x & # 178; - 2x + 3, when x ∈ [T, t + 1], find the maximum and minimum function values of the function
- 15. The following equation is: y = the absolute value of X, the absolute value of y = x, 5x & # 178; - y = 0, X & # 178; - Y & # 178; = 0. Where the number of functions where y is x is () A.0 B.1 C.2 D.4
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