If the equation x ^ 2 + 2x + k = 0 about X has real roots, then the value range of K is

If the equation x ^ 2 + 2x + k = 0 about X has real roots, then the value range of K is

(2)^2 - 4k >= 0
4-4k >= 0
k

It is known that the equation x2-2x + k = 0 about X has real roots, then the value range of K is ()
A. k＜1B. k≤1C. k≤-1D. k≥1

Δ = b2-4ac = (- 2) 2-4k = 4-4k ≥ 0, ｜ K ≤ 1

On the equation SQR (2-2x ^ 2) - KX + 2K = 0 of X, there are two unequal real roots

√(2－2x²)=k(x－2)
Namely:
√(1－x²)=(√2/2)k(x－2)
Let: Y1 = √ [(1-x & # 178;)], y2 = (√ 2 / 2) K (X-2)
The curve Y1 represents the upper half of a circle with radius 1 and center (0,0); the curve y2 = (√ 2 / 2) K (X-2) represents the straight line passing through point (2,0)
Combined with the image, the original equation should have two unequal real roots, that is, as long as the two curves have two different intersections, then:
－(√3/3)

Given that the equation M2 ^ 2x + (2m-1) 2 ^ x + M = 0 has two roots on (- ∞, 1), find the value range of M

According to the two equations, m ≠ 0,
∵ x belongs to (- ∞, 1),
∴0

The solution of equation 9.6 = 1.6x is (), 3x + 2 = 8, then 6x + 7 = ()

The first is x = 6, the second is 19

How to solve the equation 40.1x-9.6x = 6.1

40.1x-9.6x=6.1
30.5x=6.1
x=6.1/30.5
x=0.2

The equation of degree 1 with two variables, {6x + 4Y = 4.6 {3x + 1y = 1.9
If it's finished within today, 5 points will be added

6x+4y=4.6 ①
3x+1y=1.9 ②
①-2*②
2y=0.8
y=0.4
Substituting (1)
6x+1.6=4.6
6x=3
x=0.5

Square of 4x + 6x + 6 = 0

2x^2+6x+6=0
•2（x^2+3x+3）=0
•x^2+3x+3=0
•(x+1)(x+3)=0
• X = - 1 or - 3

How to solve the equation of 5x + 13 = 6x-7? If I answer it today, I'll add another 100 points
No process, I won't give you 100 points

5X+13=6X-7
5x + 13 + 7 = 6x (difference + subtraction = subtracted)
5X+20=6X
6x-5x = 20 (one addend = and - another addend)
X=20

Equation 5x + 6 = 6x-5

If 6x-5 > = 0, then x > = 5 / 6, so 5x + 6 > = 0
Remove the absolute value and get x = 11