Question

(х2+x)2-5(x2+x)+6=0​

Answer 1

Ответ:

делаем замену:

y=x^2+x

получим:

y^2-5y+6=0

D=25-24=1

y1=(5+1)/2=3

y2=(5-1)/2=2

обратная замена:

x^2+x=2

x^2+x-2=0

D=1+8=9

x1=(-1+3)/2=1

x2=-4/2=-2

x^2+x=3

x^2+x-3=0

D=1+12=13

x3=(-1+sqrt(13))/2

x4=(-1-sqrt(13))/2

Ответ: x1=1; x2=-2; x3=(-1+sqrt(13))/2; x4=(-1-sqrt(13))/2

Answer 2

It looks like you have a quadratic equation that you'd like to solve for the variable x. The equation is:

(2x + x)^2 - 5(x^2 + x) + 6 = 0

Let's simplify and solve the equation step by step:

1. Simplify the equation: (3x)^2 - 5x^2 - 5x + 6 = 0

9x^2 - 5x^2 - 5x + 6 = 0

4x^2 - 5x + 6 = 0

2. Now you have a quadratic equation in the form ax^2 + bx + c = 0, where a = 4, b = -5, and c = 6.

You can solve this quadratic equation using the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a

Plugging in the values: x = (5 ± √((-5)^2 - 4 * 4 * 6)) / (2 * 4) x = (5 ± √(25 - 96)) / 8 x = (5 ± √(-71)) / 8

Since the square root of a negative number is an imaginary number, this quadratic equation has complex solutions: x = (5 ± √71 * i) / 8

So, the solutions are: x = (5 + √71 * i) / 8 x = (5 - √71 * i) / 8

These solutions are in terms of the imaginary unit "i," indicating that they are complex numbers.