Question

) - 7x2 + 12x + 4 > 0;​

Answer 1

Ответ:

x<2

Объяснение:

-7x*2+12+4>0

-14x+12x+4>0

-2x+4>0

-2x>-4

x<2

x ∈ (-∞;2)

Answer 2

To solve the inequality $-7x^2 + 12x + 4 > 0$, we'll first find the critical points (where the expression equals zero) and then determine the intervals where the expression is greater than zero.

1. Begin by finding the critical points. These are the values of $x$ where $-7x^2 + 12x + 4 = 0$. We can solve this quadratic equation by factoring or using the quadratic formula.

Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = -7$, $b = 12$, and $c = 4$:

$x = \frac{-12 \pm \sqrt{12^2 - 4(-7)(4)}}{2(-7)}$

$x = \frac{-12 \pm \sqrt{144 + 224}}{-14}$

$x = \frac{-12 \pm \sqrt{368}}{-14}$

$x = \frac{-12 \pm 4\sqrt{23}}{-14}$

So, the critical points are $x = \frac{-12 + 4\sqrt{23}}{-14}$ and $x = \frac{-12 - 4\sqrt{23}}{-14}$.

2. Now that we have the critical points, we need to determine the intervals where the expression $-7x^2 + 12x + 4$ is greater than zero.

To do this, we can use the method of testing intervals. We'll test the three intervals defined by the critical points:

a) $x < \frac{-12 - 4\sqrt{23}}{-14}$ b) $\frac{-12 - 4\sqrt{23}}{-14} < x < \frac{-12 + 4\sqrt{23}}{-14}$ c) $x > \frac{-12 + 4\sqrt{23}}{-14}$

3. Pick a test point in each interval and evaluate the expression $-7x^2 + 12x + 4$ at that point to determine its sign in each interval.

Let's choose $x = -1$ for interval a): $-7(-1)^2 + 12(-1) + 4 = -7 - 12 + 4 = -15 + 4 = -11$ Since the result is negative, the expression is negative in interval a).

Now, choose $x = 0$ for interval b): $-7(0)^2 + 12(0) + 4 = 4$ Since the result is positive, the expression is positive in interval b).

Finally, choose $x = 1$ for interval c): $-7(1)^2 + 12(1) + 4 = -7 + 12 + 4 = 9$ Since the result is positive, the expression is positive in interval c).

4. Based on the sign of the expression in each interval:

• Interval a): $-7x^2 + 12x + 4 < 0$
• Interval b): $-7x^2 + 12x + 4 > 0$
• Interval c): $-7x^2 + 12x + 4 > 0$

So, the solution to the inequality $-7x^2 + 12x + 4 > 0$