Question

(3x+2)2=(x+2)(x-3);​

Answer 1

Ответ:

1)6x+4=x^2-3x+2x-6

x^2-7x-10=0

x1 = (7 - корень(89))/2

x2= (7 +корень(89))/2

2)x^2+2x-x-2=2x^2-4x-x+2

x^2+x=2x^2-5x+4

x^2-6x+4=0

x1=3+корень(5)

x2=3-корень(5)

3)3x^2-2x+9x-6=8x^2-12x+10x-15

5x^2-9x-9=0

x1=(9-3*корень(29))/10

x2=(9+3*корень(29))/10

Пошаговое объяснение:

Answer 2

Let's solve the given equation step by step:

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

Expanding the left side:

(9x^2 + 12x + 4) = (x^2 - x - 6)

Now let's simplify and rearrange the equation:

9x^2 + 12x + 4 = x^2 - x - 6

Bringing all terms to one side:

9x^2 + 12x - x^2 + x + 4 + 6 = 0

Combining like terms:

8x^2 + 13x + 10 = 0

At this point, we have a quadratic equation in standard form (ax^2 + bx + c = 0). To solve it, we can use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

For our equation, a = 8, b = 13, and c = 10. Substituting these values into the quadratic formula:

x = (-13 ± √(13^2 - 4 * 8 * 10)) / (2 * 8)

Simplifying further:

x = (-13 ± √(169 - 320)) / 16 x = (-13 ± √(-151)) / 16

Since the discriminant (b^2 - 4ac) is negative, we have complex roots. The square root of a negative number requires the use of imaginary numbers. Therefore, the equation does not have real solutions.

Hence, there are no real solutions to the equation (3x + 2)^2 = (x + 2)(x - 3).