For the equation of X, 4 (x + k) = 2x - (k-1), when k is equal to what value, the solution of the equation is negative

For the equation of X, 4 (x + k) = 2x - (k-1), when k is equal to what value, the solution of the equation is negative

4(X+K)=2X-(K-1)
4X+4K=2X-K+1
4X+4K-2X+K-1=0
2X+5K-1=0
2X=1-5K
X=(1-5K)/2
(1-5K)/2>0
When K1 / 5, the solution is negative

How to transform equation x ^ 2-2x + 10 = 0 into quadratic function y = a (X-H) ^ 2 + k

The main method is to use a ^ 2 ± 2Ab + B ^ 2 = (a ± b) ^ 2
y=x^2-2x+10
y=(x^2-2x+1)+9
y=(x-1)^2+9

It is known that in the expansion of (√ X-1 / (2 ^ 4 * √ x) ^ n, the absolute values of the first three coefficients form an arithmetic sequence in turn. (1) it is proved that there is no constant term in the expansion. (2) all rational terms in the expansion are solved

(√ X-1 / (2 ^ 4 * √ x) ^ n. are you sure it's the fourth power of 2, not the fourth power of X? In my experience, it should be the latter. If it's the latter, according to the binomial expansion, the absolute values of the first three terms are 1, N / 2, n (n-1) / 8 respectively. We can know that n = 1 + n (n-1) / 8 can get n = 1 (rounding off) or n = 8 from them

If two points a (- 1, - 5), B (3, - 2) are known, and the inclination angle of line L is half of that of line AB, then the slope of line L is:______ ．

Let the inclination angle of the straight line l be α, then the inclination angle of the straight line AB is 2 α, and its slope tan2 α = − 5 + 2 − 1 − 3 = 34. By using the tangent function formula of the double angle, we can get 2tan α 1 − tan2 α = 34, and simplify it to 3tan2 α + 8tan α - 3 = 0, that is, (3tan α - 1) (Tan α + 3) = 0. We can get Tan α = - 3 or tan α = 13 from the solution. If 2 α is an acute angle from tan2 α = 34 > 0, then α ∈ (0, π 4), Tan α = 13. So the answer is: 13

Find the straight line equation which passes through point P (1,2) and makes the distance from a (2,3), B (0, - 5) equal

(1) When a (2,3), B (0, - 5) are on the same side of the line, we get that the line AB is parallel to the line, K AB = 4, so the slope of the line is 4, Y-2 = 4 (x-1), which is reduced to 4x-y-2 = 0, so the line satisfying the condition is 4x-y-2 = 0, or x = 1

The equation of line L passing through point P (- 1,2) and equal to the distance between points a (2,3) and B (- 4,5)

I haven't been working on the problem for several years. I don't know if it's right? Y = - 1 / 3x + 5 / 3. If I pass through the line L with the same distance between two points of AB, I can see that the slope of line L and line AB is equal,
That is, the slope k = y1-y2 / x1-x2 = - 1 / 3
Linear equation y = KX + B = - 1 / 3x + B and passing through P point
P is substituted to get b = 5 / 3,
So the line L is y = - 1 / 3x + 5 / 3

What is the equation of a straight line passing through point P (0,1) with the same distance from a (3,3) B (5, - 1)

Let the equation of line AB be y = KX + B, then
3K + B = 3, 5K + B = - 1, k = - 2, B = 9, so y = - 2x + 9
The line passing through the point P (0,1) with the same distance from a and B is the line parallel to the line ab. therefore, K1 = k = - 2, B1 = 1,
The equation is y = - 2x + 1

If a straight line is drawn through point P (1,2) so that the distances between a (2,3) and B (4, - 5) are equal, then the equation of the straight line is ()
A. 4X + Y-11 = 0b. X + 4y-6 = 0C. 4x + Y-11 = 0 or 3x + 2y-7 = 0d. 4x + y-6 = 0 or 3x + 2y-7 = 0

When the line is parallel to AB, the slope is k = 3 + 52 − 4 = - 4, so the equation is Y-2 = - 4 (x-1), and it can be reduced to 4x + y-6 = 0

The distance between the straight line L and two parallel straight lines 2x-y + 2 = 0 and 2x-y + 4 = 0 is equal. The equation for finding the straight line L
The distance between line L and two parallel lines 2x-y + 2 = 0 and 2x-y + 4 = 0 is equal
Detailed process required

Let the linear equation 2x-y + k = 0
It is known that,
|K-2 | / √ 5 = | K-4 | / √ 5 (distance formula between parallel lines)
The solution is k = 3,
The linear equation 2x-y + 3 = 0

Find the linear equation that passes through point a (2, - 1) and the distance from point B (- 1,1) is 3

Let the equation be y + 1 = K (X-2), that is, kx-y-2k-1 = 0, and let the distance from B (- 1,1) to the line be 3, | - K − 1 − 2K − 1 | K2 + 1 = 3, | k = 512, | - line equation be y + 1 = 512 (X-2), that is, 5x-12y-22 = 0, | - line equation be x = 2 or 5x-12y-22 = 0