If (k-1) x & # 178; - radical 1-k x + quarter = 0 has two real roots, then the value range of K is?

If (k-1) x & # 178; - radical 1-k x + quarter = 0 has two real roots, then the value range of K is?

k-1≠0 ①
△>0 ②
These two conditions can be used to solve the range of K

X1, X2 are the two roots of the equation x ^ 2 + X + M = 0 about X, and | x1 | + | x2 | = 3, find the value of real number M

According to Weida's theorem, if X1 + x2 = - 1, then X1 and X2 cannot be positive numbers at the same time
If x1

Given that the two roots X1 and X2 of the equation 3x ^ 2-6 (m-1) x + m ^ 2 + 1 = 0 satisfy | x1 | + | x2 | = 2, find the value of real number M

Product of two x1 × x2 = (m ^ 2 + 1) / 3 > 0
So x1, X2 are the same number
So | x1 | + | x2 | = | X1 + x2 | = | 2 (m-1) | = 2
So m = 0 or 2
Then the discriminant = [6 (m-1)] ^ 2-12 (m ^ 2 + 1) = 24 (m ^ 2-3m + 1) > 0
M = 2, rounding off
So m = 0

Let x1, X2 be the two real numbers of the linear equation x + X + n-2 = MX with respect to x, and X1 < 0, x2-3x1 < 0, () a, M > 1, n > 2, B. m > 1, n
M.,N.

x^2+x+n-2=m
That is, x ^ 2 + X (1-m) + (n-2) = 0
Because x1, X2 are two equations
According to Weida's theorem, we get the following results
x1+x2=-(1-m)/1=m-1(1)
x1*x2=n-2
So: from (1) we get x2 = m-1-x1

It is known that two of the equations 2x & sup2; - (√ 3 + 1) x + M = 0 about X are sin α, cos α, α ∈ (0,2 π)
(1) The value of sin α / 1-cos α + cos α / 1-tan α
(2) The value of M
(3) Two solutions of the equation and the value of α

1. According to Weida's theorem, sin α + cos α = (√ 3 + 1) / 2 (1) sin α * cos α = m / 2 (2); (1) ^ 2:2sin α. Cos α = [(3 + 2 √ 3 + 1) / 4] - 1 = √ 3 / 2sin2 α = √ 3 / 2,2 α = 60 ° or 2 α = 120 ° α = 30 ° or α = 60 ° sin30 / (1-cos30) + cos30 / (1-tan30) = (1 / 2) / (1 - √

Given that the two quadratic functions x ^ 2-ax + B = 0 are sin β and cos β, the trajectory equation of P (a, b) is obtained
x=a=sinβ+cosβ(1)
y=b=sinβcosβ(2)
From (1) (2), we get x ^ 2 = 2Y + 1
The value of the answer is 0

So a ^ 2-4b = 1 + 2Sin β cos β - 4sin β cos β = 1 - 2Sin β cos β = 1 - sin2 β > = 0
The solution of β can be arbitrary
Sin β + cos β = radical 2 * (sin β / radical 2 + cos β / radical 2) = radical 2 * sin (β + 45 ')
The negative root sign 2 = 0
The solution of β can be arbitrary
To sum up, negative root 2

Given the point P (1 + cos α, sin α), where α ∈ [0, π], find the trajectory equation of point P,

x=1+cosα,y=sinα
Because (sin α) ^ 2 + (COS α) ^ 2 = 1
So (x-1) ^ 2 + y ^ 2 = 1
Because α ∈ [0, π]
So y = sin α ∈ [0,1]
So the trajectory equation of point P is (x-1) ^ 2 + y ^ 2 = 1 (Y > 0) (that is, the upper part of the circle)
If you don't understand, please hi me, I wish you a happy study!

It is known that sin α and cos α are two roots of the quadratic equation 2x2 + ax + B = 0, then the ordinary equation of the locus of point (a, b) is___ ．

We can get sin α + cos α = - 12a ①，sinα•cosα=b2… ② 1 + 2 sin α cos α = 14a2. 2 substituting into the expression, we can get: 1 + B = 14a2, that is, B = 14a2-1. The trajectory equation is: B = 14a2-1. So the answer is: B = 14a2-1

Given sin θ + cos θ = √ 3 / 2, we can solve the quadratic equation of one variable with sin θ and cos θ as roots

If a and B are real numbers and the two real number roots of the quadratic equation x ^ 2-ax + B = 0 are sin @ and COS @ (- pi / 4)

According to Weida's theorem: sin @ + cos @ = a; a = sin @ + cos @ = (radical 2) sin (@ + 45 degrees), 0