If the equation 9x + (4 + a) · 3x + 4 = 0 about X has a solution, then the value range of real number a is___ .

If the equation 9x + (4 + a) · 3x + 4 = 0 about X has a solution, then the value range of real number a is___ .

Let 3x = t > 0, then the equation 9x + (4 + a) · 3x + 4 = 0, that is T2 + (4 + a) t + 4 = 0, has positive real solution. So a = T2 + 4T + 4-T = - 4 - (T + 4T). From the basic inequality, we can get t + 4T ≥ 4. If and only if t = 4T, the equal sign holds, so - (T + 4T) ≤ - 4, so - 4 - (T + 4T) ≤ - 8, that is a ≤ - 8, so the answer is {a | a ≤ - 8}
Let X1 and X2 be the two real roots of the equation x ^ 2-2ax + A + B = 0, and find the minimum value of (x1-1) ^ 2 + (x2-1) ^ 2
X1 and X2 are the two real roots of the equation x ^ 2-2ax + A + B = 0
x1+x2=2a x1*x2=a+b
And △ = (- 2A) ^ 2-4 (a + b) ≥ 0, a ^ 2 ≥ a + B = X1 * x2
(X1-1)^2+(X2-1)^2
=(x1^2-2x1+1)+(x2^2-2x2+1)
=(x1+x2)^2-2x1*x2-2(x1+x2)+2
=4a^2-4a+2-2x1*x2
≥4a^2-4a+2-2a^2
=2(a^2-2a+1)
=2(a-1)^2
≥0
So the minimum value is 0
I hope I can help you. I wish you progress in your study_ ∩)O
(x1-1) ^ 2 + (x2-1) ^ 2 (x2-1) ^ 2 (x1-1) (x2-1) ^ 2 (x2-1) ^ 2 is (X ; + X 8322; (x 8322; (x; (x;) (x; + X;) (x; + X; (x; (x; (x; (x; (x; (x 8322;);;;; (178; (x; + X; + X; (x; (x \\\ \\\\\\\\\\becausethe equation has a solution, so (2a) &# 178; - 4 (a + b) ≥ 0, (2) a & # 178; ≥ a + B,, is expanded by (1...)
(x1-1) ^ 2 + (x2-1) ^ 2 (x2-1) ^ 2 (x1-1) (x2-1) ^ 2 (x2-1) ^ 2 is (X ; + X 8322; (x 8322; (x; (x;) (x; + X;) (x; + X; (x; (x; (x; (x; (x; (x 8322;);;;; (178; (x; + X; + X; (x; (x \\\ \\\\\\\\\\becausethe equation has a solution, so (2a) &# 178; - 4 (a + b) ≥ 0, (2) a & # 178; ≥ a + B,, from (1) (2)
The minimum value of (x1-1) ^ 2 + (x2-1) ^ 2 is 4 (a + b) - 6a-2b + 2 = 2b-2a + 2
The minimum value is 0
first
(X1-1)^2+(X2-1)^2≥0
Then (x1-1) ^ 2 + (x2-1) ^ 2
=(X & # 8321; + X & # 8322;), # 178; - 2 (X & # 8321; + X & # 8322;) - 2x & # 8321; X & # 8322; + 2, from the equation, we know X1 + x2 = 2A, X1 * x2 = a + B, and substitute it to get
The original formula = 4A & # 178; - 6a-2b + 2 has the expansion of
first
(X1-1)^2+(X2-1)^2≥0
Then (x1-1) ^ 2 + (x2-1) ^ 2
=(X & # 8321; + X & # 8322;), # 178; - 2 (X & # 8321; + X & # 8322;) - 2x & # 8321; X & # 8322; + 2, from the equation, we know X1 + x2 = 2A, X1 * x2 = a + B, and substitute it to get
The original formula = 4A & # 178; - 6a-2b + 2, and then the equation has a real number solution, we get △ = (- 2A) ^ 2-4 (a + b) ≥ 0, a ^ 2 ≥ a + B, that is, B ≤ a ^ 2-a
The original formula ≥ 4A & # 178; - 6a-2 (a ^ 2-A) + 2
=2(a^2-2a+1)
=2(a-1)^2
≥0
The minimum value is 0, where a = 1, B = 0, X1 = x2 = 1. Put it away
Zero
It is known that the quadratic equation MX ^ 2 - √ (M + 1) x + 1 = 0 with respect to X has a real root, and the value range of M is obtained
M is not equal to 0
From √ (M + 1), M + 1 > = 0 and M > = - 1 are obtained;
From the fact that the equation has real roots, the discriminant > = 0 is obtained,
Then there is [- √ (M + 1)] ^ 2-4m > = 0, and the solution is m
What is it? The root sign (M + 1) is multiplied by X
If x2 + 4y2 = (x + 2Y) 2 + a = (x-2y) 2 + B, then a and B are equal to ()
A. 4xy，4xyB. 4xy，-4xyC. -4xy，4xyD. -4xy，-4xy
∵ x2 + 4y2 = x2 + 4xy + 4y2 + a = x2-4xy + 4y2 + B, ∵ a = - 4xy, B = 4xy
In exercise 1, if the equation 2aX & # 178; - X-1 = 0 has exactly one solution in (0,1), then the value range of the real number a is? Please give the process,
When a = 0, - X-1 = 0, x = - 1 is not suitable
When a ≠ 0, △≥ 0
（-1)^2-4*2a*(-1)≥0
1+8a≥0
a≥-1/8
Let f (x) = 2aX ^ 2-x-1
There is exactly one solution in (0,1)
Then f (0) * f (1) 1
In conclusion, a > 1
Including rational number and irrational number. Among them, irrational number is infinite non cyclic decimal, rational number includes integer and fraction. Mathematically, real number is directly defined as the number corresponding to the point on the number axis. Originally, real number was only called number, but later the concept of imaginary number was introduced. The original number was called "real number" - meaning "real number".
When a = 0, x = - 1,
When a ≠ 0, the equation 2aX & # 178; - X-1 = 0 has a solution in (0,1)
Delta ≥ 0,... Expansion
Including rational number and irrational number. Among them, irrational number is infinite non cyclic decimal, rational number includes integer and fraction. Mathematically, real number is directly defined as the number corresponding to the point on the number axis. Originally, real number was only called number, but later the concept of imaginary number was introduced. The original number was called "real number" - meaning "real number".
When a = 0, x = - 1,
When a ≠ 0, the equation 2aX & # 178; - X-1 = 0 has a solution in (0,1)
Delta ≥ 0, put it away
Given that functions X1 and X2 are the two roots of equation x-2ax + A + 6 = 0, then (x1-1) + (x2-1) is the minimum?
X1 + x2 = 2A, x1x2 = a + 6 if the equation has roots (- 2A) ^ 2-4 (a + 6) ≥ 0, a ≥ 3, or a ≤ - 2. (x1-1) + (x2-1) = X1 ^ 2-2x1 + 1 + x2 ^ 2-2x2 + 1 = (x1 + x2) ^ 2-2x1x2-2 (x1 + x2) + 2 = 4A ^ 2-2a-12-4a + 2 = 4A ^ 2-6a-10, the above formula decreases on (- infinity, - 2], and obtains the minimum value 18 at - 2
If both of the quadratic equations MX ^ 2-2x + 3 = 0 are greater than - 1, then the value range of M is larger than - 1
The discriminant is greater than or equal to 0
1: When the sum of M > 0 is greater than - 2, f (- 1) > 0
2: When m
Given that the value of 2Y's Square plus y minus 2 is 3, what is the value of 4Y's Square plus 2 and y plus 1?
2y^2+y-2=3
2y^2+y=5
4y^2+2y+1
=2(2y^2+y)+1
=2*5+1
=11
Dizzy!!! The title should be solved by Baidu
It is known that the equation X3 + (1-A) x2-2ax + A2 = 0 has only one real root, then the value range of real number a is______ .
The original equation is transformed into (x-a) (x2 + x-a) = 0, and x = a or x2 + x-a = 0, because the equation X3 + (1-A) x2-2ax + A2 = 0 has and only has one real root, so x = a is the only real root of the equation, so the equation x2 + x-a = 0 has no real root, so △ = 1 + 4A < 0, so a < - 14
It is known that x1x2 is the value of | x1-x2 | of two solutions of the quadratic equation x ^ 2-4x 1 = 0
x^2-4x+1=0
x^2-4x+4=3
(x-2)^2=3
X1, X2 = 2 ± root 3
|X1-x2 | = 2 radical 3
2 × root sign 3, the specific value can be obtained