Question

(x-4)^4-5(x-4)^2-6=0

Answer 1

(х-4)²=t; t>0   
t²-5t-6=0   

{t₁+t₂=5   
{t₁*t₂=-6   

{t₁=-1 - не подходит
{t₂=6

(х-4)²=6
х²-8х+16-6=0
х²-8х+10=0
D=64-40=24
x1=(8-2√6)/2=4-√6
x2=(8+2√6)/2=4+√6

Answer 2

To solve the equation $(x-4)^4 - 5(x-4)^2 - 6 = 0$, we can use a substitution to simplify the equation. Let's introduce a new variable, $u = (x-4)^2$. Then, the equation becomes:

$u^2 - 5u - 6 = 0$.

Now, we can solve this quadratic equation for $u$:

$u^2 - 5u - 6 = 0$.

To factor this quadratic equation, we look for two numbers whose product is -6 and whose sum is -5.

The two numbers are -6 and +1, as (-6) * (+1) = -6 and (-6) + (+1) = -5.

So, we can rewrite the equation as:

$(u - 6)(u + 1) = 0$.

Now, set each factor equal to zero and solve for $u$:

1. $u - 6 = 0$: $u = 6$.

2. $u + 1 = 0$: $u = -1$.

Now, we have two values for $u$, but we still need to find the corresponding values for $x$. Recall that $u = (x-4)^2$.

1. For $u = 6$: $(x-4)^2 = 6$. Taking the square root of both sides: $x - 4 = ±√6$. Solving for $x$: $x = 4 ± √6$.

2. For $u = -1$: $(x-4)^2 = -1$. Since the square of any real number cannot be negative, this equation has no real solutions.

Thus, the solutions to the original equation are $x = 4 + √6$ and $x = 4 - √6$.