Question

25^\sqrt{x} - 124 * 5^\sqrt{x} = 125

Answer 1

25^(√х) - 124•5^(√х) = 125

Ограничение: х ≥ 0

Пусть 5^(√х) = а , а > 0 , тогда

а² - 124а - 125 = 0

По теореме Виета находим следующие корни:

а₁ = 125 ⇔ 5^(√х) = 125  ⇔  5^(√х) = 5³  ⇔  √х = 3  ⇔  х = 9

а₂ = - 1 - не подходит

ОТВЕТ: 9

Answer 2

$25^{\sqrt{x}}-124\cdot 5^{\sqrt{x}}=125\; \; ,\; \; \; \; ODZ:\; x\geq 0\\\\t=5^{\sqrt{x}}>0\; \; ,\; \; t^2-124t-125=0\; \; ,\; \; t_1=-1\; ,\; t_2=125\; \; (teorema\; Vieta)\\\\t_1=-1$

Answer 3

To solve the equation $25^{\sqrt{x}} - 124 \cdot 5^{\sqrt{x}} = 125$, we can use a substitution to simplify it. Let's make a substitution:

Let $y = 5^{\sqrt{x}}$. Then the equation becomes:

$25^{\sqrt{x}} - 124 \cdot 5^{\sqrt{x}} = 125$ $y^2 - 124y - 125 = 0$

Now we have a quadratic equation in terms of $y$. We can solve this quadratic equation using factoring or the quadratic formula:

$y^2 - 124y - 125 = 0$ $(y - 125)(y + 1) = 0$

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

1. $y - 125 = 0 \implies y = 125$
2. $y + 1 = 0 \implies y = -1$

Since $y = 5^{\sqrt{x}}$, we need to solve for $x$ in each case:

1. $y = 125 \implies 5^{\sqrt{x}} = 125$

We know that $125 = 5^3$, so:

$5^{\sqrt{x}} = 5^3$ $\sqrt{x} = 3$

Now, square both sides to get rid of the square root:

$x = 3^2$ $x = 9$

2. $y = -1$

Since $y$ is the base raised to a power, it cannot be negative. Therefore, this case has no valid solutions.

So, the only valid solution for the equation $25^{\sqrt{x}} - 124 \cdot 5^{\sqrt{x}} = 125$ is $x = 9$.