Question

(10x-9)^4-19(10x-9)^2-92=0

Answer 1

(10x-9)⁴-19*(10x-9)-92=0

Пусть (10x-9)²=t>0      ⇒

t²-19t-92=0      D=729         √D=27

t₁=-4 ∉           t₂=23       ⇒

(10x-9)²=23

100x²-180x+81=23

100x²-180x+58=0   |÷2

50x²-90x+29=0    D=2300      √D=10*√23

Ответ: x₁=0,9+(√23)/10           x₂=0,9-(√23)/10.

Answer 2

(10x - 9)^2 = t.
t^2 - 19t - 92 = 0.
t1 = -4 t2 = 23
(10x - 9)^2 = -4 - нету корней.
(10х - 9)^2 = 23
1)10 х - 9 = -√23

$x = \frac{ - \sqrt{23} + 9 }{10}$
2) 10 х - 9 = √23
$x = \frac{ \sqrt{23} + 9}{10}$

Answer 3

To solve the equation $(10x - 9)^4 - 19(10x - 9)^2 - 92 = 0$, we can use a substitution to make it easier to handle. Let's define a new variable $y$ such that $y = (10x - 9)^2$. Now, the equation becomes:

$y^2 - 19y - 92 = 0$

This is a quadratic equation in terms of $y$, and we can solve it using factoring or the quadratic formula. Factoring this quadratic equation, we have:

$(y - 23)(y + 4) = 0$

Setting each factor to zero and solving for $y$:

1. $y - 23 = 0 \implies y = 23$
2. $y + 4 = 0 \implies y = -4$

Now that we have two potential values for $y$, we can substitute back to find the corresponding values for $x$:

1. $y = (10x - 9)^2 = 23$ $(10x - 9)^2 = 23$ $\sqrt{(10x - 9)^2} = \sqrt{23}$ $10x - 9 = \pm \sqrt{23}$ $10x = 9 \pm \sqrt{23}$ $x = \frac{9 \pm \sqrt{23}}{10}$

2. $y = (10x - 9)^2 = -4$ Since $y$ represents the square of a real number, there are no real solutions for this case.

Therefore, the solutions to the original equation are:

$x = \frac{9 + \sqrt{23}}{10} \quad \text{and} \quad x = \frac{9 - \sqrt{23}}{10}$