Question

A) 1+ (m+2)² =3(2m-1)

Answer 1

Ответ:

розв'язку немає

Объяснение:

Д меньше 0

Answer 2

To solve the equation $1 + (m + 2)^2 = 3(2m - 1)$, follow these steps:

Step 1: Expand the squared term on the left side. $1 + (m + 2)^2 = 1 + (m + 2)(m + 2) = 1 + (m^2 + 2m + 2m + 4) = 1 + (m^2 + 4m + 4) = m^2 + 4m + 5$

Step 2: Distribute the 3 on the right side of the equation. $3(2m - 1) = 3 * 2m - 3 * 1 = 6m - 3$

Now, the equation becomes: $m^2 + 4m + 5 = 6m - 3$

Step 3: Move all terms to one side to set the equation to zero. $m^2 + 4m - 6m + 5 + 3 = 0$

Step 4: Combine like terms. $m^2 - 2m + 8 = 0$

Now, you have a quadratic equation in the form $ax^2 + bx + c = 0$, where $a = 1$, $b = -2$, and $c = 8$.

Step 5: Solve the quadratic equation using either factoring, completing the square, or the quadratic formula.

The quadratic equation does not factor easily, so let's use the quadratic formula to find the solutions for $m$:

The quadratic formula states: $m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Plugging in the values: $m = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(8)}}{2(1)}$

$m = \frac{2 \pm \sqrt{4 - 32}}{2}$

$m = \frac{2 \pm \sqrt{-28}}{2}$

Since the expression under the square root ($-28$) is negative, the solutions will involve complex numbers. We can simplify it as follows:

$m = \frac{2 \pm i \sqrt{28}}{2}$

$m = 1 \pm i\sqrt{7}$

So, the two solutions for $m$ are $m = 1 + i\sqrt{7}$ and $m = 1 - i\sqrt{7}$. These are complex numbers.