Question

Решите 4x^2+4x+1⩾0 .

Answer 1

Ответ(-0.5;+∞)

Пошаговое объяснение:

D=16-4*4*1=16-16=0

x=-4:8=-1/2

(x+0.5)>0

(-0.5;+∞)

Answer 2

To solve the inequality $4x^2 + 4x + 1 \geq 0$, we can follow these steps:

Step 1: Factor the quadratic expression if possible. Step 2: Find the critical points by setting the expression equal to zero. Step 3: Determine the sign of the expression in each interval created by the critical points. Step 4: Write the solution to the inequality.

Step 1: Factor the quadratic expression: The expression $4x^2 + 4x + 1$ cannot be factored further. Therefore, we proceed with the next steps.

Step 2: Find the critical points by setting the expression equal to zero: $4x^2 + 4x + 1 = 0$

To solve this quadratic equation, we can use the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

For our equation, $a = 4$, $b = 4$, and $c = 1$:

$x = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 4 \cdot 1}}{2 \cdot 4}$

$x = \frac{-4 \pm \sqrt{16 - 16}}{8}$

$x = \frac{-4}{8} = -\frac{1}{2}$

So, the only critical point is $x = -\frac{1}{2}$.

Step 3: Determine the sign of the expression in each interval created by the critical points:

We have one critical point at $x = -\frac{1}{2}$, which divides the number line into two intervals:

Interval 1: $x < -\frac{1}{2}$ Interval 2: $x > -\frac{1}{2}$

To check the sign of the expression in each interval, we can pick a test point from each interval and substitute it into the original inequality. We'll choose $x = 0$ for Interval 1 and $x = 1$ for Interval 2:

For $x = 0$: $4x^2 + 4x + 1 = 4(0)^2 + 4(0) + 1 = 1$ Since $1 \geq 0$, the expression is positive or zero in Interval 1.

For $x = 1$: $4x^2 + 4x + 1 = 4(1)^2 + 4(1) + 1 = 9$ Since $9 \geq 0$, the expression is positive or zero in Interval 2.

Step 4: Write the solution to the inequality:

The inequality $4x^2 + 4x + 1 \geq 0$ is satisfied for all values of $x$ in both intervals:

$x \in (-\infty, -\frac{1}{2}] \cup (-\frac{1}{2}, \infty)$

So, the solution to the inequality is $-\infty < x \leq -\frac{1}{2}$ or $x > -\frac{1}{2}$.