Question

X² - 2x +√(2 - x) = √(2 - x) + 3​

Answer 1

Ответ:

х1=3, х2=-1

Пошаговое объяснение:

корнивычеркиваем, получается х2-2х=3

х2-2х-3=0 по теореме Виета корни уравнения 3, -1

Answer 2

x² - 2x +√(2 - x) = √(2 - x) + 3​

ОДЗ: x≤2, т.к. если x будет равен 3 и более, то придётся извлекать корень из отрицательного числа.

√(2 - x) взаимоуничтожается. Получаем выражение

x²-2x-3=0

По теореме Виета (устно)

x₁=-1

x₂=3 - не подходит по ОДЗ

Ответ: x=-1

Answer 3

To solve the equation $x^2 - 2x + \sqrt{2 - x} = \sqrt{2 - x} + 3$, we can start by isolating the terms with square roots on one side of the equation and moving all other terms to the opposite side. Here's how you can do it step by step:

1. Subtract $\sqrt{2 - x}$ from both sides of the equation: $x^2 - 2x + \sqrt{2 - x} - \sqrt{2 - x} = 3$

2. The square root terms on the left side will cancel each other out, leaving: $x^2 - 2x = 3$

3. Now, move the constant term (3) to the left side by subtracting 3 from both sides: $x^2 - 2x - 3 = 0$

4. You now have a quadratic equation in standard form. To solve it, you can use the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

   In this case, $a = 1$, $b = -2$, and $c = -3$. Plug these values into the quadratic formula:

   $x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-3)}}{2(1)}$

   Simplify:

   $x = \frac{2 \pm \sqrt{4 + 12}}{2}$ $x = \frac{2 \pm \sqrt{16}}{2}$ $x = \frac{2 \pm 4}{2}$

5. Now, you have two possible solutions:

   a. $x = \frac{2 + 4}{2} = \frac{6}{2} = 3$

   b. $x = \frac{2 - 4}{2} = \frac{-2}{2} = -1$

So, the solutions to the equation $x^2 - 2x + \sqrt{2 - x} = \sqrt{2 - x} + 3$ are $x = 3$ and $x = -1$.