Given that the equation AX2 + C = 0 (where a is not equal to 0) has no real root, the sign relation between a and C is The roots of the quadratic equation AX2 + BX + C = 0 (a is not equal to 0) are determined by the values of what, what and what in the equation

Given that the equation AX2 + C = 0 (where a is not equal to 0) has no real root, the sign relation between a and C is The roots of the quadratic equation AX2 + BX + C = 0 (a is not equal to 0) are determined by the values of what, what and what in the equation

A and C have the same sign. The values of a, B and C are determined by discriminant, two sum and product formula

Given a ^ 2 + B ^ 2-2 (a + 2b) = - 5, find the solution of the equation ABX ^ 2 + BX + ax + a = 0

(a ^ 2 - 2A + 1) + (b ^ 2 - 4B + 4) = 0; (A-1) ^ 2 + (B - 2) ^ 2 = 0; a = 1; b = 2; so ABX ^ 2 + BX + ax + a = 0, that is: 2x ^ 2 + 2x + X + 1 = 0; 2x ^ 2 + 3x + 1 = 0; (x + 1) (2x + 1) = 0; X1 = - 1; x2 = - 1 / 2; x2 = - 1 / 2; x2 = - 1 / 2;

Given the equation (AX + 1) &# 178; = A & # 178; (1-x) &# 178;, where a > 1, it is proved that the positive root of the equation is smaller than 1 and the negative root is larger than - 1
If the answer is complete, the reward will be increased!

Equation (AX + 1) &# 178; = A & # 178; (1-x & # 178;)
It can be reduced to f (x) = 2 (AX) &# 178; + 2aX + 1-A & # 178;
When a > 1,
f(-1)=(a-1)²>0,
f(0)=(1+a)(1-a)0,
So the roots X1 and X2 of F (x) = 0 satisfy - 1

On the equation AX & # 178; - 2 (a + 1) x + A-1 = 0 (a > 0), when we find the value of a, the equation has one root, two positive roots and two negative roots
On the equation AX & # 178; - 2 (a + 1) x + A-1 = 0 of X
(a > 0) what is the value of a
The equation has one root
Two positive roots
Two negative roots
One positive and one negative
One is greater than one and the other is less than one
Both are greater than 1
One is between (1,2) and the other is between (3,4)
It's all between [1,2]
Only one is between (1,2)

Analysis: let two be x1, x2
① Ax2-2 (a + 1) x + A-1 = 0 has one positive and one negative
Then △ = 4 (a + 1) ^ 2-4a (A-1) = 12a + 4 > 0,
A ＞ - 1 / 3
In addition, X1 * x2 = (A-1) / a < 0,
The solution is 0 < a < 1
∴0＜a＜1
② Ax2-2 (a + 1) x + A-1 = 0, both are greater than 1,
Then △ = 4 (a + 1) ^ 2-4a (A-1) = 12a + 4 ≥ 0,
x1+x2=2(a+1)/a,x1*x2=(a-1)/a
(x1-1)+(x2-1)=x1+x2-2=2(a+1)/a-2＞0
(x1-1)*(x2-1)=x1*x2-(x1+x2)+1
=(a-1)/a-2(a+1)/a+1＞0
Solution a ≥ - 1 / 3
a＞0
a＜0
In this case, a has no solution, that is, there is no a with both equations greater than 1

·(X/4)^2+[(56-X)/4]^2=196

X = 56 or x = 0

27.1x2 + 4x = 80.6 is solved by equation

27.1x2+4x=80.6 54.2+4x=80.6
4x=80.6-54.2
4x=26.4
x=6.6

How to solve the equation 27-4x = 4.6

27-4x=4.6
4x=27-4.6
4x=22.4
x=22.4/4
x=5.6

How to solve this equation 4x-x = 27

3X=27
X=27/3
X=9

4X ^ 6 - 4x ^ 4 + 3 = 0 solve the equation

Let X & #178; = M
4m³ - 4m² + 3 = 0
The solution is m = - 0.67
Therefore, there is no real number solution to this problem

X-9 / 4x + 140 = 4 / 3x, how to solve this equation?

140=4/3x+9/4x-x
140=16/12x+27/12x-x
140=43/12x-x
140=43/12x-12/12x
140=21/12x
140*12/21=12/21*21/12x
x=80