Write a necessary and sufficient condition for the equation AX2 + 2x + 1 = 0 to have at least one real root Prove your conclusion

Write a necessary and sufficient condition for the equation AX2 + 2x + 1 = 0 to have at least one real root Prove your conclusion

When a = 0, 2x + 1 = 0, x = - 1 / 2 has a real root
2. When a is not equal to 0, the discriminant = 4-4a > = 0
a

There is a point P on the ellipse x2 / A2 + Y2 / B2 = 1 (a > b > 0), so that OP ⊥ AP (o is the origin, a is the endpoint of the long axis). The range of the eccentricity of the ellipse is determined by the slope

Let p be the intersection of the ellipse and the circle (x ± A / 2) ^ 2 + y ^ 2 = (A / 2) ^ 2. Let's take the circle as (x-a / 2) ^ 2 + y ^ 2 = (A / 2) ^ 2. Let's eliminate y from the simultaneous equations of the ellipse and the circle to get C ^ 2 / A ^ 2 * x ^ 2-ax + B ^ 2 = 0. The discriminant is △ = a ^ 2-4b ^ 2C ^ 2 / A ^ 2 = ((a ^ 2-2c ^ 2

2 * 4 and 899 * 5 * 25 out of 1000 calculated in a simple way! Hope to have an answer in 3 minutes

=2*4*5*25+2*899/1000*5*25
=2 * 5 * 4 * 25 + (9-1 / 100) * 25 = 1000 + 250-25-1 / 4 = 1224 and 3 / 4

If the equation kx-10x + 3 = 0 has a solution of x = 3 / 4, then k =?

Substituting x = 3 / 4 into the equation kx-10x + 3 = 0, we get that
3k/4-10×3/4+3=0
3k/4-15/2+3=0
3k/4=15/2-3
3k/4=9/2
k=9/2÷(3/4)
k=6

What is the distance from a point on the ellipse to the focus 2?
Ellipse 2c.2b 2a is an arithmetic sequence. What is eccentricity

(1) The distance from a point on the ellipse to the focus 2 is equal to 2A
(2)
2c.2b 2a is an arithmetic sequence, and C + a = 2b is obtained
There is also a "= B" + C in the ellipse“
So: 5C = 3A
Eccentricity e = C / a = 3 / 5

17 / 18:17 * 2:3-10:3 / 5:3

17 / 18:17 * 2:3-10:3 / 5:3
=18*3/2-1/2
=27-1/2
=26.5

Solve the equation. 0.4x minus four times 51 equals 60.4

35cm = (fraction) m, the simplest fraction is 40ml = (fraction) l, 538ml (fraction) cubic decimeter
28:00 = (fraction of a day)
750 kg = (fraction) ton

35 cm = (7 / 20) M
40 ml = (1 / 25) l
538 ml (269 parts per 100) cubic decimeter
28:00 = (7 / 6) days
750 kg = (3 / 4) t

What is the 2004 digit after the decimal point of the formula 12 + 13 + 14 + 15 + 16 + 17 + 18 + 19 + 110 + 111 + 112 + 113) × 2004?

2004 can be divided by 2, 3, 4, 6, 12, so 12, 13, 14, 161, 122, 004 divided by 5, 8, 10 are finite decimal, so 15, 181, 102, 0047 = 286. 285, 714, 285, 714 , is a 6-digit cycle, 2004 after the decimal point, 2005 is 4220049 = 222.66

There are several ways to solve hyperbolic standard equation

(1) Let the hyperbolic equation be x ^ 2 / A ^ 2 - y ^ 2 / b ^ 2 = 1 (a > 0, b > 0). According to the meaning of 2B = 12, B = 6, B ^ 2 = 36, e ^ 2 = C ^ 2 / A ^ 2 = (a ^ 2 + B ^ 2) / A ^ 2 = (a ^ 2 + 36) / A ^ 2 = 25 / 16, a ^ 2 = 64, the hyperbolic equation be x ^ 2 / 64 - y ^ 2 / 36 = 1 (2)