Please help me to analyze my doubts in the following problems, exercise: the equation lg2x / LG (x + a) = 2, ask why a is worth, the equation has a solution? arrangement: lg2x=2lg(x+a) 2x=(x+a)^2 The results are as follows x^2+2(a-1)x+a^2=0 And 2x > 0, x + a > 0, For the above univariate quadratic equation, △ = 4 [(A-1) ^ 2] - 4 (a ^ 2) = - 8A + 4, There are three cases ① When △ 0, - 8A + 4 > 0, a < 1 / 2 In this case, the equation has two solutions, X = {2-2a ± [radical (4-8a)} / 2 = 1-A ± [radical (1-2a)] In this case, x = (1-A) + [radical (1-2a)] > 0 is obviously true (positive number plus positive number); For x = (1-A) - [radical (1-2a)], since (1-A) ^ 2 - (1-2a) = 1-2a + A ^ 2-1 + 2A = a ^ 2 > 0, x = 1-A - [radical (1-2a)] > 0 also holds However, because x + a > 0 is required, Therefore, when a ＜ 1 / 2 and X + a ＞ 0, the original equation has two solutions ② When △ = 0, a = 1 / 2 In this case, the equation is x ^ 2-x + 1 / 4 = 0, and the unique solution is x = 1 / 2 But it is meaningless that the denominator is 0 Therefore, when x = 1 / 2, the original equation has no solution ③ When △ 0, a ＞ 1 / 2, the original equation has no solution To sum up, (1) When a < 1 / 2, the equation has two solutions; (2) There is no a such that the equation has a solution; (3) When a ≥ 1 / 2, the equation has no solution My question is: 1, "for x = (1-A) - [root (1-2a)], since (1-A) ^ 2 - (1-2a) = 1-2a + A ^ 2-1 + 2A = a ^ 2 > 0", how can I get 1-A - √ (1-2a) > 0 from this? 2. "When a ＜ 1 / 2 and X + a ＞ 0, the original equation has two solutions." why? From 2x > 0, we can get x > 0; X + a > 0, we can get x > - A. why not find the maximum value of - x, and then combine A0 and 2x > 0, that is, x > 0 and x > - A. should we also find the maximum value of - x, and then find the range of a? Even if we can't find it, should we also meet a > - x? Why only a > 2

Please help me to analyze my doubts in the following problems, exercise: the equation lg2x / LG (x + a) = 2, ask why a is worth, the equation has a solution? arrangement: lg2x=2lg(x+a) 2x=(x+a)^2 The results are as follows x^2+2(a-1)x+a^2=0 And 2x > 0, x + a > 0, For the above univariate quadratic equation, △ = 4 [(A-1) ^ 2] - 4 (a ^ 2) = - 8A + 4, There are three cases ① When △ 0, - 8A + 4 > 0, a < 1 / 2 In this case, the equation has two solutions, X = {2-2a ± [radical (4-8a)} / 2 = 1-A ± [radical (1-2a)] In this case, x = (1-A) + [radical (1-2a)] > 0 is obviously true (positive number plus positive number); For x = (1-A) - [radical (1-2a)], since (1-A) ^ 2 - (1-2a) = 1-2a + A ^ 2-1 + 2A = a ^ 2 > 0, x = 1-A - [radical (1-2a)] > 0 also holds However, because x + a > 0 is required, Therefore, when a ＜ 1 / 2 and X + a ＞ 0, the original equation has two solutions ② When △ = 0, a = 1 / 2 In this case, the equation is x ^ 2-x + 1 / 4 = 0, and the unique solution is x = 1 / 2 But it is meaningless that the denominator is 0 Therefore, when x = 1 / 2, the original equation has no solution ③ When △ 0, a ＞ 1 / 2, the original equation has no solution To sum up, (1) When a < 1 / 2, the equation has two solutions; (2) There is no a such that the equation has a solution; (3) When a ≥ 1 / 2, the equation has no solution My question is: 1, "for x = (1-A) - [root (1-2a)], since (1-A) ^ 2 - (1-2a) = 1-2a + A ^ 2-1 + 2A = a ^ 2 > 0", how can I get 1-A - √ (1-2a) > 0 from this? 2. "When a ＜ 1 / 2 and X + a ＞ 0, the original equation has two solutions." why? From 2x > 0, we can get x > 0; X + a > 0, we can get x > - A. why not find the maximum value of - x, and then combine A0 and 2x > 0, that is, x > 0 and x > - A. should we also find the maximum value of - x, and then find the range of a? Even if we can't find it, should we also meet a > - x? Why only a > 2

As for problem 1, only one step was omitted when solving the problem. The following inference is just the product of two solutions. Because the first solution has been determined to be positive, the product is proved, and the second solution is proved, and vice versa

The motion equation of the object is s (T) = 4T ^ 3 + 2T ^ 2 + 3T, and the acceleration of the object at t = 2 is calculated
It is helpful for the responder to give an accurate answer

The derivative of displacement with respect to time is v = DS / dt (D is the infinitesimal element, a small section), while acceleration is the derivative of velocity with respect to time, expressed as a = DV / dt. Therefore, the expression of acceleration is to continuously calculate the second derivative of displacement
The problem is s' (T) = 12t ^ 2 + 4T + 3
s''(t)=24t+4
So when t = 2, the acceleration is 52 m / s Square

The equation of motion of the object is s = - 13t3 + 2t2-5, then the instantaneous velocity of the object at t = 3 is______ ．

When t = 3, the instantaneous velocity of the object is - 32 + 4 × 3 = 3, so the answer is 3