Question

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

Answer 1

Сделаем замену:
(5x+6)^2=t,t>0
t^2+5t-6=0
t=-6-не удовлетворяет
   t=1-подставим в замену:
(5x+6)^2=1
|5x+6|=1

5x+6=1              5x+6=-1
5x=-5                    5x=-7
x=-1                         x=-7/5

Answer 2

$(5x+6)^4 + 5 (5x+6)^2 - 6 = 0$ ;

Обозначим $y = (5x+6)^2 ,$                                       формула [1]

тогда $y^2 = ((5x+6)^2)^2 = (5x+6)^4$ ;                    формула [2]


Заменяя в исходном уравнении скобки
на выражения в формулах [1] и [2], получаем

$y^2 + 5 y - 6 = 0$ ;

Дискриминант: $D = 5^2 - 4 \cdot 1 \cdot (-6) = 25 + 24 = 49 = 7^2$ ;

$y_1 = \frac{-b-\sqrt{D}}{2a} = \frac{ -5 -\sqrt{7^2} }{ 2 \cdot 1 } = \frac{ -5 -7 }{2} = \frac{ -12 }{2} = -6$ ;

$y_2 = \frac{-b+\sqrt{D}}{2a} = \frac{ -5 +\sqrt{7^2} }{ 2 \cdot 1 } = \frac{ -5 + 7 }{2} = \frac{ 2 }{2} = 1$ ;


Возвращаемся к $x$ через формулу [1] :

$\left[\begin{array}{ll} -6 = (5x+6)^2 ; & impossible \\ 1 = (5x+6)^2 . & \end{array}\right$

$\left[\begin{array}{l} x \in \emptyset ; \\ \left[\begin{array}{l} 5x+6 = -1 ; \\ 5x+6 = 1 . \end{array}\right \end{array}\right$

$\left[\begin{array}{l} 5x = -1-6 ; \\ 5x = 1-6 . \end{array}\right$

$\left[\begin{array}{l} 5x = -7 ; \\ 5x = -5 . \end{array}\right$

$\left[\begin{array}{l} x = -7:5 ; \\ x = -5:5 . \end{array}\right$

$\left\begin{array}{l} x = -1.4 ; \\ x = -1 . \end{array}\right$



О т в е т : $x \in \{ -1.4 , -1 \} .$

Answer 3

To solve the given equation $(5x+6)^4 + 5(5x+6)^2 - 6 = 0$, you can use substitution to simplify the equation. Let $y = (5x + 6)^2$. Now the equation becomes a quadratic equation in terms of $y$:

$y^2 + 5y - 6 = 0$

Factor the quadratic equation:

$(y + 6)(y - 1) = 0$

This gives two possible values for $y$:

1. $y + 6 = 0 \Rightarrow y = -6$
2. $y - 1 = 0 \Rightarrow y = 1$

Now substitute back $y = (5x + 6)^2$ into these equations and solve for $x$:

1. For $y = -6$: $(5x + 6)^2 = -6$ There are no real solutions for this case.

2. For $y = 1$: $(5x + 6)^2 = 1$ Taking the square root of both sides, you get: $5x + 6 = \pm 1$ Solve for $x$ in each case:

   • $5x + 6 = 1$ $5x = -5$ $x = -1$
   • $5x + 6 = -1$ $5x = -7$ $x = -\frac{7}{5}$

Therefore, the solutions to the given equation are $x = -1$ and $x = -\frac{7}{5}$.