Question

1) (x+1) (x+2)-(x+3) (x+4)=0 2) (2xв квадрате-(4х-3) (3х+1)= -2

Answer 1

1) (x+1) (x+2)-(x+3) (x+4)=0
x^2+2x+x+2-x^2-4x-3x-12=0
-4x-10=0
-4x=10
x=-2.5
2) (2x^2-(4х-3) (3х+1)= -2
2x^2-12x^2-4x+9x+3=-2
-10x^2+5x+5=0 разделим все на -5
2x^2-x-1=0
D=b^2-4ac=1+8=9
x1=-b-√D/2a=1-3/2=-1
x2=-b+√D/2a=1+3/2=2

Answer 2

Let's solve the two equations you've provided:

1. (x+1)(x+2) - (x+3)(x+4) = 0

First, expand the expressions:

(x^2 + 3x + 2) - (x^2 + 7x + 12) = 0

Now, simplify the equation:

x^2 + 3x + 2 - (x^2 + 7x + 12) = 0

Now, distribute the negative sign in front of the second parenthesis:

x^2 + 3x + 2 - x^2 - 7x - 12 = 0

Now, simplify further by combining like terms:

(3x - 7x) + (2 - 12) = 0

-4x - 10 = 0

Add 10 to both sides:

-4x = 10

Now, divide by -4:

x = 10 / -4

x = -5/2

So, the solution to the first equation is x = -5/2.

2. (2x^2 - (4x-3)(3x+1)) = -2

First, expand the expressions:

2x^2 - (12x^2 - x - 4x + 3) = -2

Now, simplify the equation:

2x^2 - 12x^2 + x + 4x - 3 = -2

Combine like terms:

(2x^2 - 12x^2) + (x + 4x) - 3 = -2

-10x^2 + 5x - 3 = -2

Now, add 2 to both sides:

-10x^2 + 5x - 3 + 2 = 0

-10x^2 + 5x - 1 = 0

Now, divide the entire equation by -1 to make it easier to work with:

10x^2 - 5x + 1 = 0

Now, you can use the quadratic formula to solve for x:

x = (-b ± √(b^2 - 4ac)) / (2a)

In this case, a = 10, b = -5, and c = 1. Plugging these values into the quadratic formula:

x = (-(-5) ± √((-5)^2 - 4 * 10 * 1)) / (2 * 10)

x = (5 ± √(25 - 40)) / 20

x = (5 ± √(-15)) / 20

Since the discriminant (the value inside the square root) is negative, there are no real solutions to this equation. It means that there are no real values of x that satisfy the second equation.