Question

1) (х+3)^4-3(х+3)^2-4=0 2)(6х-7)^4+4(6х-7)^2+3=0

Answer 1

1) Замена: (x+3)^2=t, тогда

t^2 -3t -4=0

D=9-4*(-4)=25

t1= -1

t2= 4

(x+3)^2= -1

Не может быть

(x+3)^2= 4

x+3= 2   x+3= -2

x= -1.      x= -5

2) Замена: (6x-7)^2=t, тогда

t^2 + 4t +3 =0

t1*t2= 3.     t1= -3

t1+t2= -4    t2= -1

(6x-7)^2= -3

Не может быть.

(6x-7)^2= -1

Нельзя.

(?)

Answer 2

To solve these equations, we can use a substitution method. Let's substitute a new variable to simplify the expressions.

For the first equation, let's substitute u = (x + 3). Now we have:

u^4 - 3u^2 - 4 = 0

This is a quadratic equation in terms of u^2. Let's solve it:

u^4 - 3u^2 - 4 = 0

Let's substitute v = u^2:

v^2 - 3v - 4 = 0

Factoring the equation, we have:

(v - 4)(v + 1) = 0

Setting each factor to zero, we get:

v - 4 = 0 or v + 1 = 0

Solving these equations, we have:

v = 4 or v = -1

Now, substitute back u = u^2:

u^2 = 4 or u^2 = -1

For the first equation, u^2 = 4, we have two possibilities:

u = 2 or u = -2

Substituting back x + 3 = u:

x + 3 = 2 or x + 3 = -2

For x + 3 = 2, we have:

x = 2 - 3 = -1

For x + 3 = -2, we have:

x = -2 - 3 = -5

So the solutions for the first equation are x = -1 and x = -5.

For the second equation, let's follow the same process. Let's substitute w = 6x - 7:

w^4 + 4w^2 + 3 = 0

This equation is a quadratic equation in terms of w^2. Let's solve it:

w^4 + 4w^2 + 3 = 0

Substituting z = w^2:

z^2 + 4z + 3 = 0

Factoring the equation, we have:

(z + 3)(z + 1) = 0

Setting each factor to zero, we get:

z + 3 = 0 or z + 1 = 0

Solving these equations, we have:

z = -3 or z = -1

Now, substitute back w = w^2:

w^2 = -3 or w^2 = -1

For w^2 = -3, we have no real solutions.

For w^2 = -1, we have:

w = i or w = -i

Substituting back 6x - 7 = w:

6x - 7 = i or 6x - 7 = -i

For 6x - 7 = i, we have:

6x = 7 + i

x = (7 + i)/6

For 6x - 7 = -i, we have:

6x = 7 - i

x = (7 - i)/6

So the solutions for the second equation are x = (7 + i)/6 and x = (7 - i)/6.