Question

(3x-1)(x+4)=x в квадрате -3x-4

Answer 1

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

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

3x^2+12x-x-4-x^2+3x+4=0

2x^2+14x=0

x (2x+14)=0

x=0 2x+14=0

2x= -14

x= -7

Ответ: х=0, х= -7.

Answer 2

Let's begin by expanding the left-hand side of the equation using the distributive property:

(3x-1)(x+4) = 3x^2 + 12x - x - 4

Simplifying the expression, we get:

(3x-1)(x+4) = 3x^2 + 11x - 4

Now we can compare this to the right-hand side of the equation:

x^2 - 3x - 4

We see that the left-hand side and the right-hand side are not equal. Therefore, the given equation is not true for all values of x.

Let's solve for x by setting the left-hand side equal to the right-hand side and simplifying:

3x^2 + 11x - 4 = x^2 - 3x - 4

2x^2 + 14x = 0

2x(x + 7) = 0

x = 0 or x = -7

However, we need to check whether these values satisfy the original equation or not:

When x = 0, the left-hand side of the equation is (-1)(4) = -4, and the right-hand side is 0^2 - 3(0) - 4 = -4. So x = 0 satisfies the equation.

When x = -7, the left-hand side of the equation is (3(-7) - 1)(-7+4) = (-22)(-3) = 66, and the right-hand side is (-7)^2 - 3(-7) - 4 = 49 + 21 - 4 = 66. So x = -7 also satisfies the equation.

Therefore, the solutions to the equation are x = 0 and x = -7.