Solution equation: (1 / 2 times x + 2) (3x-1) - 3 / 2 times the square of X + 13 = 0

Solution equation: (1 / 2 times x + 2) (3x-1) - 3 / 2 times the square of X + 13 = 0

(1/2x+2)(3x-1)-3/2x²+13=0
3/2x²+11/2x-2-3/2x²+13=0
11/2x+11=0
11/2x=-11
11x=-22
x=-2

The absolute value of x-3 is greater than or equal to 4 solutions

|x-3|≥4
Then: x-3 ≥ 4 3-x ≥ 4
x≥7 x≤-1

It is known that x is less than 0, y is greater than 0, and the absolute value of X

X0 and | x|

Why does the equation 4x + 2n / (x + m) = x have two roots with equal absolute values and opposite signs?
I'd like to help you

(4x+2n)/(x+m)=x(x+m)/(x+m)
(xx+mx－4x－2n)/(x+m)=0
(xx+mx－4x－2n)=0
xx+(m－4)x－2n=0
There are two roots with equal absolute values and opposite signs
xx+(m－4)(-x)－2n=0
m－4=0
m=4
(xx－2n)/(x－4)=0
m=4
n> 0

Why does the equation 4x + 2n / (x + m) have two roots with equal absolute values and opposite signs?
I can't do it
Help me
It's a process

Rewrite it into the general formula ax ^ 2 + BX + C = 0
Its root is the sum of two - B / a = 0, the product of two C / A

If the two different signs are positive or negative, the absolute value is large? X ^ 2 - root sign 3 times X-5 = 0, x ^ 2-2 √ 6 * x + √ 3 = 0

X ^ 2-radical 3 times X-5 = 0
X1*X2=-5/1=-50
So the absolute value of positive root is large
X^2-2√6*X+√3=0
X1*X2=√3/1=√3>0
So the two have the same number
X1+X2=2√6>0
So it's a positive root

The absolute value of solving equation 3x-2 - 4 = 0

|3x-2|-4=0
|3x-2|=4
3x-2 = 4 or 3x-2 = - 4
X = 2 or x = - 2 / 3

For solving the equation, the two following signs of 2x & # 178; + 3x-1 = 0 are
We don't need a function image to do it

Because C / a = - 1 / 2

1. The two symbols of 2x & # 178; + 3x-1 = 0 are () A. the same sign B. the different sign C. both of them are positive D
2. If the two roots of the univariate quadratic equation x & # 178; + (M & # 178; - 1) x + M = 0 are opposite to each other, then ()
A. M = ± 1 B.M = - 1 C.M = 1 D, none of the above conclusions are correct
3. A quadratic equation of one variable with the sum of two real roots may be ()
A.x²+2x-3=0 B.x²-2x+3=0 C.x²+2x+3=0 D.x²-2x-3=0
4. If the lengths of the two sides of a triangle are 3 and 4, and the root of the third side's rectangular path X & # 178; - 12x + 35 = 0, then the perimeter of the triangle is ()
A. 14 b, 12 C, 12 or 14 d. none of the above is true
The two symbols of 1.2x & # 178; + 3x-1 = 0 are () A. same sign B. different sign C. both of them are positive D
2. If the two roots of the univariate quadratic equation x & # 178; + (M & # 178; - 1) x + M = 0 are opposite to each other, then ()
A. M = ± 1 B.M = - 1 C.M = 1 none of the above conclusions are correct
3. A quadratic equation of one variable with the sum of two real roots may be ()
A.x²+2x-3=0 x²-2x+3=0 x²+2x+3=0 x²-2x-3=0
4. If the lengths of the two sides of a triangle are 3 and 4, and the root of the third side's rectangular path X & # 178; - 12x + 35 = 0, then the perimeter of the triangle is ()
A. 14 b, 12 C, 12 or 14 d. none of the above is true

1. The image of function y = 2x & # 178; + 3x-1 passes through the point (0, - 1), and the opening is upward, so B
2. If the axis of symmetry of the function is x = - B / (2a) = 0, then M & # 178; - 1 = 0, M = ± 1; Δ = - 4m ＞ = 0, then M = - 1
3. If Δ＞ = 0, we can see that B and C do not meet the conditions and exclude them; the sum of the two real roots is - B / a = 2, and the D term x & # 178; - 2x-3 = 0 meets the conditions
4. Solve the equation and get roots 5,7. Each side of the triangle is larger than the third side, 7 = 4 + 3, excluding, so the third side is 5 in length and 12 in circumference

Solving the equation, judging the two symbols of 2x square + 3x-7 = 0

The sum of the two is less than 0, so there is a negative number. The product of the two is less than 0, so one is positive and one is negative, and the absolute value of the negative number is larger