Question

(1+x)*<x+2

Answer 1

$(1+x \sqrt{3}) \cdot\sqrt{2}$

$\sqrt{2}+x \sqrt{6}$

$x( \sqrt{6}-1)$

$x< \frac{2 \sqrt{3}- \sqrt{2} }{ \sqrt{6}-1 }$

Answer 2

Let's solve the inequality:

(1 + x) * (x + 2) < x + 2

First, let's simplify the equation:

(x + 1)(x + 2) < x + 2

Expanding the left side of the inequality:

x^2 + 3x + 2 < x + 2

Rearranging the terms:

x^2 + 3x + 2 - x - 2 < 0

Combining like terms:

x^2 + 2x < 0

Now, let's solve for x by factoring the quadratic expression:

x(x + 2) < 0

The inequality holds true when either:

1. x < 0 and x + 2 > 0, or
2. x > 0 and x + 2 < 0

Let's solve each case separately:

Case 1: x < 0 and x + 2 > 0

For x < 0, we have x + 2 > 0:

x < 0 x + 2 > 0

From the second inequality, subtracting 2 from both sides:

x > -2

So the solution for case 1 is -2 < x < 0.

Case 2: x > 0 and x + 2 < 0

For x > 0, we have x + 2 < 0:

x > 0 x + 2 < 0

From the second inequality, subtracting 2 from both sides:

x < -2

However, this contradicts the assumption that x > 0, so there are no solutions for case 2.

Therefore, the solution to the inequality (1 + x) * (x + 2) < x + 2 is -2 < x < 0.