Question

X^2+16x+64>0 как решить?

Answer 1

Для начала решим дискриминант, найдём корень уравнения и дальше на координатной оси x найдём по методу интервалов решение неравенства:

Решение:

Дискриминант: x²+16x+64 = 0

a=1 b=16 c=64

D = b²-4ac => 16²-4*1*64 = 256 - 256 = 0 равен 0, 1 корень.

x = $\frac{-b}{2a}$

x₁ = $\frac{-16}{2} = -8$

Получаем: (x+8)>0

Значит метод интервалов нам не пригодится в данном случае, ибо одно значение.

------------------------------------0///////////////////////////////////> x

-∞                                     -8                                         +∞

Ответ: (-8;+∞).

Answer 2

To solve the inequality $x^2 + 16x + 64 > 0$, we can use the concept of quadratic inequalities. First, let's find the critical points where the expression $x^2 + 16x + 64$ equals zero. These critical points will divide the number line into intervals, and we can test points from each interval to determine where the expression is positive.

Step 1: Find the critical points (where $x^2 + 16x + 64 = 0$).

The quadratic expression can be factored as $(x + 8)^2$. So, set it equal to zero and solve for $x$:

$(x + 8)^2 = 0$

Taking the square root of both sides:

$x + 8 = 0$

$x = -8$

Step 2: Analyze the intervals.

Now, we have one critical point, $x = -8$. This critical point divides the number line into two intervals: $(-∞, -8)$ and $(-8, +∞)$.

Step 3: Test points in each interval.

Pick a test point from each interval and substitute it into the inequality to determine the sign of the expression.

For $-9$: $(-9)^2 + 16(-9) + 64 = 81 - 144 + 64 = 1 < 0$ (negative)

For $-7$: $(-7)^2 + 16(-7) + 64 = 49 - 112 + 64 = 1 > 0$ (positive)

Step 4: Write the solution.

Based on the test points, we found that the expression $x^2 + 16x + 64$ is positive for $x \in (-8, +∞)$. So the solution to the inequality is:

$x \in (-8, +∞)$

In interval notation, the solution is $(-8, +∞)$. This means any value of $x$ greater than -8 will satisfy the inequality.