Exercises and answers for solving quadratic equation of one variable x*x+2x=36

x²+2x-36=0
X = {- 2 ± root sign [2 & sup2; - 4 × 1 × (- 36)]} / (2 × 1)
=(- 2 ± 2 root 37) / 2 = - 1 ± root 37
X1 = - 1 + root 37, X2 = - 1 - root 37

In the same plane rectangular coordinate system, if the intersection of the line y = 3x-1 and the line y = x-k is in the fourth quadrant, then the value range of K is ()
A. K ＜ 13b. 13 ＜ K ＜ 1C. K ＞ 1D. K ＞ 1 or K ＜ 13

The solution of the system of equations y = 3x − 1y = x − k about X and Y is: x = 1 − k2y = 1 − 3k2 ∵ the intersection point is in the fourth quadrant ∵ the system of inequalities 1 − K2 > 01 − 3k2 < 0 is obtained; the solution is 13 ∵ K < 1, so B is selected

In the same plane rectangular coordinate system, if the intersection of the line y = 3x-1 and the line y = x-k is in the fourth quadrant, then the value range of K is ()
A. K ＜ 13b. 13 ＜ K ＜ 1C. K ＞ 1D. K ＞ 1 or K ＜ 13

In the plane rectangular coordinate system, it is known that the intersection of the line x-2y = - K + 1 and the line x + 3Y = 4K + 6 is in the fourth quadrant, and the value of integer k is obtained

Solve the equation x − 2Y = − K + 1x + 3Y = 4K + 6 about X, y, the solution is: x = K + 3Y = K + 1, ∵ the intersection point is in the fourth quadrant ∵ get the inequality system K + 3 > 0k + 1 < 0, the solution is - 3 < K < - 1, ∵ K is a positive integer, ∵ K is - 2. Answer: the value of integer k is - 2

In the same plane rectangular coordinate system, if the intersection of the line y = 3x-1 and the line y = x-k is in the fourth image line
Why subtract

Subtraction is to find out the abscissa m of the focus after simultaneous establishment of two linear equations, and then bring back any equation to get the ordinate n of the intersection point, and the intersection point is in the fourth quadrant, so m > 0, n

If the intersection of the first-order function y = 2x + 5m-1 and y = - 1 / 3x + m-2 is in the fourth quadrant, the value of M is obtained

First, we find out the relationship between X = * * m and y = * * * M
Then make x > 0 y

Given that the intersection point of the line y = - 2X-4 and the line y = 3x + B is in the second quadrant, the value range of B is obtained
Because. So

-2x-4=3x+b
-5x=b+4
x=-(b+4)/5
y=-2*[-(b+4)/5]-4
=(2b+8)/5-4
=(2b-12)/5
The intersection is in the second quadrant: x0
-(b+4)/50
b>-4
(2b-12)/5>0
b-6>0
b>6
So the value range of B is b > 6

Given that the intersection points of the lines y = 1 / 2x + K / 2-3 and y = - 1 / 3x + 4 / 3K + 1 / 3 are in the fourth quadrant, what is the value range of K? If K is a non negative integer, △ Pao is an isosceles triangle with AO as the base, the coordinates of point a are (2,0), and point P is on the line y = 1 / 2x + K / 2-3, the coordinates of point P are obtained

Solution: (1) 1 / 2x + K / 2-3 = - 1 / 3x + 4 / 3K + 1 / 3
x=k+4
y=1/2x+k/2-3=k-1
So the intersection coordinates of the two lines are (K + 4, k-1)
∵ focus on the fourth quadrant
∴k+4>0,k-1

Given that the intersection point of the line y = 2x + 1 and the line y = 3x + B is in the second quadrant, the value range of B is obtained

y=2x+1=3x+b
x=1-b
y=2x+1=3-2b
The second quadrant x = 1-b1
y=3-2b>0,b

Find the extremum of function f (x) = 13x3 − 4x + 4

∵f(x)＝13x3−4x+4，∴f′（x）=x2-4=（x-2）（x+2）． … If f '(x) = 0, the solution is x = 2, or x = - 2 When f '(x) > 0, i.e. x > 2, or x < 2; when f' (x) < 0, i.e. - 2 < x < 2. When x changes, the changes of F '(x), f (x) are as follows: X (- ∞, - 2) - 2 (- 2, 2) 2 (2, + ∞) f' (x) & nbsp; + 0_ 0 + F (x) monotonically increasing 283 & nbsp; monotonically decreasing − 43 & nbsp; monotonically increasing Therefore, when x = - 2, f (x) has a maximum, and the maximum is f (- 2) = 283; when x = 2, f (x) has a minimum, and the minimum is f (2) = -43 12 points

Exercises and answers for solving quadratic equation of one variable x*x+2x=36