Question

(x-3)^4-5(x-3)^2-14=0

Answer 1

Сделаем замену: пусть (x-3)² = t, тогда:
t² -5t - 14 = 0
D = 25 + 56 = 81 = 9²
t₁ =  $\frac{5 + 9}{2} = 7$
t₂ = $\frac{5 - 9}{2} = -2$

1) (x - 3)² = 7
x² - 6x + 9 - 7 = 0
x² - 6x + 2 = 0
D = 36 - 8 = 28 = 2√7
x₁ = 3 + √7
x₂ = 3 - √7

2) (x - 3)² = -2
x² - 6x + 9 + 2 = 0
x² - 6x + 11 = 0
D = 36 - 44 = ∅

Answer 2

To solve the equation $(x-3)^4 - 5(x-3)^2 - 14 = 0$, we can use a substitution to simplify the equation. Let $u = (x-3)^2$. Then, the equation becomes:

$u^2 - 5u - 14 = 0$.

Now, we have a quadratic equation in terms of $u$. We can factor it or use the quadratic formula to solve for $u$:

$u^2 - 5u - 14 = (u - 7)(u + 2) = 0$.

This gives us two possible values for $u$: $u = 7$ or $u = -2$.

Now, we'll substitute back $u = (x-3)^2$:

For $u = 7$: $(x - 3)^2 = 7$, $x - 3 = \pm \sqrt{7}$, $x = 3 \pm \sqrt{7}$.

For $u = -2$: $(x - 3)^2 = -2$. This equation has no real solutions because the square of a real number cannot be negative.

Therefore, the real solutions to the original equation $(x-3)^4 - 5(x-3)^2 - 14 = 0$ are $x = 3 + \sqrt{7}$ and $x = 3 - \sqrt{7}$.