The quadratic equation of one variable a ^ 2-2ka + 1 / 2K ^ 2-2 = 0 is known It is known that the quadratic equation of one variable a ^ 2-2ka + 1 / 2K ^ 2-2 = 0 for A. (1) it is proved that no matter what the value of K is, the equation always has two unequal real roots; (2) let a and B be two of the equations, and a ^ 2-2ka + 2Ab = 5, the value of K is obtained

（1）
△=4k^2-4*(1/2k-2)=2k^2+8>0
So the equation must have two unequal roots
（2）
A is the solution of the equation
a^2-2ka+1/2k^2-2=0
And a ^ 2-2ka + 2Ab = 5
According to Weida's theorem, ab = 1 / 2K ^ 2-2,
By substituting the above two formulas, a ^ 2-2ka and ab are eliminated
So K ^ 2 = 14
K = + - radical 14

Given that X05 + m1x + N1 = 0 and X05 + M2X + N2 = 0, and m1m2 = 2 (N1 + N2), try to prove that at least one of the two equations has a real root

The value of the discriminant of two equations
△1=m1²-4n1,△2=m2²-4n2
So △ 1 + △ 2 = M1 & # 178; + M2 & # 178; - 4 (N1 + N2) = M1 & # 178; + M2 & # 178; - 2m1m2 = (m1-m2) & # 178; > = 0
So at least one of the two discriminant is greater than or equal to 0,
So at least one of the two equations has a real root

It is known that formula A: x2 + P1x + Q1 = 0, formula B: x2 + P2X + Q2 = 0, where P1, P2, Q1 and Q2 are all real numbers and satisfy p1p2 = 2 (Q1 + Q2)
Ask whether at least one of the two equations has a real root, and explain the reason

X^2+P1X+Q1=0;X^2+P2X+Q2=0
Sum of two discriminant equations
Δ1+Δ2
=（p1^2-4q1）+(p2^2-4q2)
=p1^2+p2^2-2(2q1+q2)
=p1^2+p2^2-2p1p2
=(p1-p2)^2
≥ 0
So Δ 1, Δ 2 must have one ≥ 0
That is: at least one equation has real roots

If X & # 178; + P1x + Q1 = 0 and X & # 178; + P2X + Q2 = 0, it is proved that when p1p2 = 2 (Q1 + Q2), at least one of the two equations has a real root

The discriminant of the former equation root is △ 1 = P1 ^ 2-4q1, and the discriminant of the latter equation root is △ 2 = P2 ^ 2-4q2 ∧ 1 + Δ 2 = P1 ^ 2-4q1 + P2 ^ 2-4q2 = P1 ^ 2 + P2 ^ 2-4q1-4q2 = (P1-P2) ^ 2 + 2 (p1p2-2p1-2p2). When p1p2 = 2p1 + 2P2, p1p2-2p1-2p2 = 0 ∧

It is proved that the equation (m ^ 2 + 1) x ^ 2 - (M + 2) x + 3 = 0 has no real root

b^2-4ac=(m+2)^2-4*3*(m^2+1) =m^2+4m+4-12m^2-12 =-11m^2+4m-8 =-11[m^2-(4/11)m+4/121]+(4/121-8) =-11(m-2/11)^2-964/121

It is proved that the equation (x + 1) (x + 3) = A-3 must have two unequal real roots

(x + 1) (x + 3) = A-3, that is, x ^ 2 + 4x + 3 = a ^ 2-3, x ^ 2 + 4x + 6-A ^ 2 = 0, discriminant = 4 ^ 2-4 * 1 * (6-A ^ 2) = 16-4 (6-A ^ 2) = - 8 + a ^ 2, so when discriminant = - 8 + A ^ 2 > 0, that is, a ^ 2 > 8, there must be two unequal real roots

If the value of y = MX ^ 2-mx + 1 is always positive, then the value range of real number m is?

Meet the conditions of m m > 0 and &; = m ^ 2-4m (m-2) 8 / 3,

If the line MX + 2Y + 1 = 0 and X + Y-2 = 0 are perpendicular to each other, then the real number M=______ ．

∵ the straight line MX + 2Y + 1 = 0 and the straight line x + Y-2 = 0 are perpendicular to each other, the product of the slopes is equal to - 1, and the solution is m = - 2, so the answer is: - 2

Let the line mx-y + 2 = 0 be tangent to the circle x2 + y2 = 1 & nbsp;, then the value of the real number m is ()
A. 3b. - 3C. 3 or - 3D. 2

∵ circle x2 + y2 = 1, ∵ center of circle (0, 0), radius r = 1, and straight line mx-y + 2 = 0 is tangent to circle, ∵ distance from center of circle to straight line d = R, that is, 2M2 + 1 = 1, the solution is: M = ± 3, then the value of real number m is 3 or - 3

The straight line MX + Y - M = O, no matter what real number m takes, it passes through the fixed point P

MX + y-m = 0, that is, m (x-1) + y = 0,
∵ over point P
Ψ X-1 = 0 and y = 0
The straight line passes through the fixed point P (1,0)