X (x + 1) 2x to solve quadratic equation with one variable

X (x + 1) 2x to solve quadratic equation with one variable

Solution 2x (x) = 1
That is x (x + 1) - 2x = 0
That is, X (x + 1-2) = 0
That is, X (x-1) = 0
That is, x = 0 or x = 1
（1）x2=64                      （2）5x2-25=0                 （3）（x+5）2=16（4）8（3-x）2-72=0              （5）2y=3y2（6）2（2x-1）-x（1-2x）=0（7）3x（x+2）=5（x+2）（8）（1-3y）2+2（3y-1）=0．
(1) ∵ (± 8) 2 = 64, ∵ x = ± 8, that is, X1 = 8, X2 = - 8; (2) by shifting the term, 5x2 = 25, the coefficient is 1, X2 = 225, x = ± 25, that is, X1 = 25, X2 = - 25; (3) by shifting the term, the coefficient is 1, (3-x) 2 = 9, 3-x = ± 3, that is, X1 = 6, X2 = 0; (5) by shifting the term, 3y2-2y = 0, y (3y-2) = 0, ∵ y = 0, 3y-2 = 0, the solution is Y1 = 0, y2 = 23; (6) 2x-1 ）(2 + x) = 0, 2x-1 = 0, 2 + x = 0, the solution is X1 = 12, X2 = - 2; (7) by changing the term, 3x (x + 2) - 5 (x + 2) = 0, (x + 2) (3x-5) = 0, ｜ x + 2 = 0, 3x-5 = 0, the solution is X1 = - 2, X2 = 53; (8) (3y-1) (3y-1 + 2) = 0, ｜ 3y-1 = 0, 3Y + 1 = 0, the solution is Y1 = 13, y2 = - 13
Solving quadratic equation with one variable x ^ 2 + 2x + 1 = 0
Is it better to use matching method or formula method?
If the collocation method is used, there is only one real root, while the formula method has two equal real roots
How to distinguish?
Match method, formula method will do
The matching method also has two real roots
It's just 0 on the right side of the question, which is quite special
Is two equal real roots
3x2+2x-2=3（x+ ___ ） 2+ ___ .
3x2+2x-2=3（x2+23x）-2=3（x2+23x+19-19）-2=3（x+13）2-73．
X + 2Y = Y-X / 4 = 2x + 1 / 3 to solve quadratic equation of two variables·
x+2y=y-x/4①
y-x/4=2x+1/3②
①：4x+8y=4y-x
5x+4y=0③
②：12y-3x=24x+4
12y-3x=24x+4
12y-27x=4④
③×3-④：42x=-4
x=-2/21
Substituting ③: 5 × (- 2 / 21) + 4Y = 0
4y=10/21
y=5/42
x+2y=y-x/4=2x+1/3，
It is reduced to {y = - 5x / 4,
{-3x/2=2x+1/3,
The solution is {x = - 2 / 21,
Can you write in more detail
(within 30 hours) the following is a table of the corresponding values of the independent variable x and the function value y of the quadratic function y = ax ^ 2 + BX + C (a is not equal to 0)
x …… -3,-2,-1,0,1,2,3,……
y…… 12,5,0,-3,-4,-3,0,……
Answer the following questions based on the information provided in the table above
(1) Find the coordinates of the intersection of the function image and the Y axis
(2) Is the symmetry axis of the function image on the right or left of the y-axis? Explain the reason
(3) Find the two intersection points of function image and X-axis as a and B respectively, and the vertex is C. find the area of triangle ABC
(1) （0,-3）
(2) The axis of symmetry x = 1, so it's on the right side of Y
(3) Intersection with Y-axis (- 1,0) (3,0), vertex (1, - 4)
S△abc=3-（-1）×4=16
Establish a rectangular coordinate system, trace the point, find the axis of symmetry, can also solve the equation of quadratic function, then it is simple
If the integer x.y satisfies the inequality x ^ + y ^ + 1 and is less than or equal to 2x + 2Y, then the value of X + y has
x²+y²+1≤2x+2y
(x²-2x+1)+(y²-2y+1)≤1
(x-1)²+(y-1)²≤1
Because (x-1) &# 178; ≥ 0, (Y-1) &# 178; ≥ 0, and X and y are integers, there are two cases
① (x-1) & # 178; and (Y-1) & # 178; both are 0, namely:
(x-1)²=0
(y-1)²=0
The solution is as follows
X=1
Y=1
Then: x + y = 2;
② If one of (x-1) &# 178; and (Y-1) &# 178; is 0 and the other is 1, let (x-1) &# 178; = 0, then (Y-1) &# 178; = 1,
The solution is as follows
X=1
Y-1 = ± 1, y = 2 and 0,
Then x + y = 3 and 1
To sum up, there are three values of X + Y: 1, 2 and 3
x+y=1,2,3
Given the quadratic function y = X2 - (m-1) x-2m-3, where m is a real number (1), it is proved that for any real number m, the quadratic function must have two properties
Discriminant
=(-(m-1))^2-4*(-2m-3)
=m^2-2m+1+8m+12
=m^2-6m+9+4
=(m-3)^2+4>0
Therefore, there are two intersections between the quadratic function and the X axis, that is, there are two equations
If x and y are positive numbers, then the minimum value of (x + 12Y) 2 + (y + 12x) 2 is ()
A. 3B. 72C. 4D. 92
∵ x, y are positive numbers, ∵ (x + 12Y) 2 + (y + 12x) 2 ≥ 2 (XY + 14xy + 1), the condition of equal sign is x + 12Y = y + 12x, the solution is x = y, the condition of equal sign is xy = 14xy, the condition of equal sign is xy = 14xy, the solution is x = y = 22, that is, when x = y = 22, the minimum value of (x + 12Y) 2 + (y + 12x) 2 is 4, so C should be selected
Given that the partial corresponding values of the independent variable x and the function value Y in the quadratic function y = ax ^ 2 + BX + C (a is not equal to 0) are shown in the following table: then the analytic expression of the quadratic function is______
The corresponding values of the independent variable x and the function value Y in the quadratic function y = ax ^ 2 + BX + C (a is not equal to 0) are given in the following table: then the analytic expression of the quadratic function is______