Question

(x+1)^2(x^2+2x)=12 (x+1)^4+4(x+1)^2-21=0

Answer 1

Ответ:

Для первого уравнения решением являются два корня х=1 и х=-3

Для второго уравнения равнения решением является один корень х=3/4

Пошаговое объяснение:

Для первого уравнения:

(х+1)*2(2х+2х)=12

(х+1)*4х+4х=12

$4x^{2} +4x+4x=12\\4x^{2} +8x-12=0\\a=4 ; b=8 ; c=-12\\D=8^{2}-4*4*(-12)=256\\ \sqrt{D} =\sqrt{256} =16\\x=1\\x=-3$

Для второго уравнения:

4х+4+8х+8-21=0

12х-9=0

х=9/12

х=3/4

Answer 2

To solve the equations, let's work through them one by one:

1. (x+1)^2(x^2+2x) = 12

Step 1: Expand the left-hand side (LHS) of the equation: (x+1)^2(x^2+2x) = (x^2 + 2x)(x^2 + 2x) = x^2(x^2 + 2x) + 2x(x^2 + 2x) = x^4 + 2x^3 + 2x^3 + 4x^2 = x^4 + 4x^3 + 4x^2

Step 2: Set the expanded equation equal to 12: x^4 + 4x^3 + 4x^2 = 12

Step 3: Rearrange the equation to make it equal to zero: x^4 + 4x^3 + 4x^2 - 12 = 0

Now, let's move on to the second equation:

2. (x+1)^4 + 4(x+1)^2 - 21 = 0

Step 1: Expand the left-hand side (LHS) of the equation: (x+1)^4 = (x+1)^2 * (x+1)^2 = (x^2 + 2x + 1) * (x^2 + 2x + 1) = x^4 + 2x^3 + x^2 + 2x^3 + 4x^2 + 2x + x^2 + 2x + 1 = x^4 + 4x^3 + 6x^2 + 4x + 1

Step 2: Replace the (x+1)^4 term in the equation with its expanded form: x^4 + 4x^3 + 6x^2 + 4x + 1 + 4(x+1)^2 - 21 = 0

Step 3: Replace (x+1)^2 with its expanded form: x^4 + 4x^3 + 6x^2 + 4x + 1 + 4(x^2 + 2x + 1) - 21 = 0

Step 4: Distribute the 4 inside the parentheses: x^4 + 4x^3 + 6x^2 + 4x + 1 + 4x^2 + 8x + 4 - 21 = 0

Step 5: Combine like terms: x^4 + 4x^3 + 10x^2 + 12x - 16 = 0

Now you have two polynomial equations:

1. x^4 + 4x^3 + 4x^2 - 12 = 0
2. x^4 + 4x^3 + 10x^2 + 12x - 16 = 0

To solve these equations, you can use numerical methods like graphing, factoring, or using numerical solvers. Unfortunately, these equations are not easily factorable, so I recommend using a numerical solver or graphing calculator to find the approximate solutions for x.