Question

(2х+1)²+(х+3)(х-3)=1​

Answer 1

Ответ:

1

Объяснение:

4x²+4x+1+x²-9=1

5x²+4x=9

x = 1 (метод подбора)

для справки:

(a+b)² = a²+2ab+b²

(a+b)(a-b) = a²-b²

Answer 2

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

1. Expand the expressions: $(2x+1)^2 = 4x^2 + 4x + 1$ $(x+3)(x-3) = x^2 - 9$

2. Substitute the expanded expressions back into the original equation: $4x^2 + 4x + 1 + x^2 - 9 = 1$

3. Combine like terms: $5x^2 + 4x - 8 = 1$

4. Move all terms to one side of the equation to set it equal to zero: $5x^2 + 4x - 9 = 0$

Now you have a quadratic equation in standard form. To solve for $x$, you can either factor it or use the quadratic formula. In this case, the equation is not easily factorable, so let's use the quadratic formula:

The quadratic formula is given by: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

In your equation, $a = 5$, $b = 4$, and $c = -9$. Plugging these values into the formula:

$x = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 5 \cdot -9}}{2 \cdot 5}$

Simplifying the expression under the square root: $x = \frac{-4 \pm \sqrt{16 + 180}}{10}$ $x = \frac{-4 \pm \sqrt{196}}{10}$ $x = \frac{-4 \pm 14}{10}$

So, the solutions for $x$ are: $x = \frac{10}{10} = 1$ $x = \frac{-18}{10} = -\frac{9}{5}$

Therefore, the solutions for the equation are $x = 1$ and $x = -\frac{9}{5}$.