If the two roots of the equation lg2x + (lg5 + lg7) lgx + lg5lg7 = 0 are X1 and X2, then X1 * x2 =?

If the two roots of the equation lg2x + (lg5 + lg7) lgx + lg5lg7 = 0 are X1 and X2, then X1 * x2 =?

X1 and X2 are two roots of the equation lg2x + (lg5 + lg7) lgx + lg5lg7 = 0
So there are:
lgx1+lgx2=(lg5+lg7)
lgx1lgx2=lg5*lg7
therefore
lg(x1x2)=lg(35)
therefore
x1x2=35
Note:
The equation takes lgx as the unknown!
The corresponding solution is as follows
lgx

When the equation lg2x / LG (x + a) = 2, what is the value of a, 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, because (1-A) ^ 2 - (1-2a) = 1-2a + A ^ 2-1 + 2A = a ^ 2 > 0, how to get x = 1-A - [radical (1-2a)] > 0 is also true
2. "So, 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, and a > - X. why not find the maximum value of - x, and then combine A0 and 2x > 0, that is, x > 0, a > - X,
Should we also find the maximum value of - x, and then find the value range of A. even if we can't find it, should we also meet a > - x? Why only a > 1 / 2?

lg2x=2lg(x+a)
2x=(x+a)^2
x^2+2(a-1)x+a^2=0 --- (1)
△=4[(a-1)^2]-4(a^2)=-8a+4=4(1-2a)
The original equation requires x > 0 and X + a > 0, and X + A is not equal to 1
So,
When 1-2a1 / 2, equation (1) has no solution, then: the original equation has no solution
When 1-2a = 0, i.e. a = 1 / 2, equation (1) has a solution x = 1-A = 1 / 2
But at this time, x + a = 1, so that the original equation does not hold
Therefore, when a = 1 / 2, the original equation has no solution
When 1-2a > 0, that is A0
X2 = 1-a-radical (1-2a) = (1 / 2) [radical (1-2a) - 1] ^ 2 > = 0
But: X1 + a = 1 + radical (1-2a) > 1
So, X1 must be a real root of the original equation
When the root sign (1-2a) - 1 = 0, that is, a = 0, X2 = 0 is not the root of the original equation, so at this time, the original equation has only one real root
When a is not equal to 0, then x2 > 0,
And X2 + a = 1-radical (1-2a) 0, then: X2 is also the root of the original equation, then: the original equation has two real roots
At this point, 1-radical (1-2a) > 0, a > 0, combined with a

Exercise: please help me to analyze my doubts in the following problems. The equation lg2x / LG (x + a) = 2. When I ask what a is, 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

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