Question

X^2<16 решит неравинства

Answer 1

2²<16
3²<16
но это может быть не правильно

Answer 2

To solve the inequality $x^2 < 16$, we can break it down into two separate inequalities and find the solution for each part. Remember that when solving inequalities involving quadratics, we often start by factoring the expression if possible.

1. $x^2 - 16 < 0$: Factor the left side as a difference of squares. $(x + 4)(x - 4) < 0$

2. $x^2 - 16 = (x + 4)(x - 4)$

Now, we need to find the intervals where the expression $(x + 4)(x - 4)$ is less than zero (i.e., negative).

To do that, we can use the concept of signs in intervals.

1. Set $(x + 4)(x - 4) = 0$ to find critical points: $x = -4$ and $x = 4$.

2. Plot these critical points on a number line:

   (-∞)---(-4)---(4)---(∞)

3. Choose test points in each interval:

   • Test $x = -5$ in $(x + 4)(x - 4): (-5 + 4)(-5 - 4) = (-1)(-9) = 9$ [Positive]
   • Test $x = 0$ in $(x + 4)(x - 4): (0 + 4)(0 - 4) = (4)(-4) = -16$ [Negative]
   • Test $x = 5$ in $(x + 4)(x - 4): (5 + 4)(5 - 4) = (9)(1) = 9$ [Positive]

4. Determine the sign in each interval:

   (-∞)---(-4)---(4)---(∞)

   • (−) (−) (+)

The sign chart tells us that the expression $(x + 4)(x - 4)$ is negative in the interval $-4 < x < 4$.

So, the solution to the inequality $x^2 < 16$ is $x$ belongs to the open interval $-4 < x < 4$, or in interval notation: $(-4, 4)$.