Question

(2+x)(2-x)<(0,5x+1)(3-2x)

Answer 1

$(2+x)(2-x)\ \textless \ (0,5x+1)(3-2x)\\\\4-x^2\ \textless \ 1,5x-x^2+3-2x\\\\1\ \textless \ -0,5x\\\\x$

Answer 2

To solve the inequality $(2+x)(2-x) < (0.5x+1)(3-2x)$, you can start by simplifying both sides of the inequality and then finding the solution set for $x$.

Let's simplify each side step by step:

1. Expand the left side: $(2+x)(2-x) = 4 - x^2$

2. Expand the right side: $(0.5x+1)(3-2x) = 1.5x - x^2 + 3 - 2x$

Now, we have the inequality: $4 - x^2 < 1.5x - x^2 + 3 - 2x$

Next, we can combine like terms on the right side: $4 - x^2 < -0.5x + 3$

Now, let's isolate the variable (x:

1. Add $x^2$ to both sides: $4 < -0.5x + x^2 + 3$

2. Subtract 3 from both sides: $1 < -0.5x + x^2$

3. Rearrange the terms: $x^2 - 0.5x - 1 < 0$

Now, you have a quadratic inequality. To solve it, you can either use the quadratic formula or factor it. Factoring is a bit tricky in this case, so let's use the quadratic formula:

The quadratic formula is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ for the quadratic equation $ax^2 + bx + c = 0$.

In our case, $a = 1$, $b = -0.5$, and $c = -1$.

Plug these values into the formula:

$x = \frac{-(-0.5) \pm \sqrt{(-0.5)^2 - 4(1)(-1)}}{2(1)}$

Simplify:

$x = \frac{0.5 \pm \sqrt{0.25 + 4}}{2}$

$x = \frac{0.5 \pm \sqrt{4.25}}{2}$

Now, you have two possible solutions:

1. $x = \frac{0.5 + \sqrt{4.25}}{2}$
2. $x = \frac{0.5 - \sqrt{4.25}}{2}$

You should also check the discriminant ($b^2 - 4ac$) to determine the nature of the solutions:

$0.25 - 4(1)(-1) = 0.25 + 4 = 4.25$

Since the discriminant is positive, you have two real solutions.

So, the solution set for the inequality $(2+x)(2-x) < (0.5x+1)(3-2x)$ is:

$x \in \left(\frac{0.5 - \sqrt{4.25}}{2}, \frac{0.5 + \sqrt{4.25}}{2}\right)$