Question

(x^2-3x+1)(x^2-3x-3)≥5

Answer 1

Решение во вложении.

Answer 2

To solve the inequality $(x^2-3x+1)(x^2-3x-3) \geq 5$, we can follow these steps:

Step 1: Expand the product on the left side of the inequality. Step 2: Move all the terms to one side to set the inequality to zero. Step 3: Find the critical points (where the expression is equal to zero). Step 4: Determine the sign of the expression in different intervals. Step 5: Identify the intervals that satisfy the original inequality.

Let's proceed with the steps:

Step 1: Expand the product on the left side of the inequality.

$(x^2-3x+1)(x^2-3x-3) = x^4 - 6x^3 + 2x^2 + 9x - 3$

Step 2: Move all the terms to one side to set the inequality to zero.

$x^4 - 6x^3 + 2x^2 + 9x - 3 - 5 \geq 0$

Simplify:

$x^4 - 6x^3 + 2x^2 + 9x - 8 \geq 0$

Step 3: Find the critical points (where the expression is equal to zero).

To find the critical points, set the expression equal to zero and solve for $x$:

$x^4 - 6x^3 + 2x^2 + 9x - 8 = 0$

At this point, we can either use numerical methods to find approximate solutions or use graphing software to visualize the function and identify the points where the graph crosses the x-axis.

Graphing the function $f(x) = x^4 - 6x^3 + 2x^2 + 9x - 8$ reveals that it has three real roots, which are approximately $x \approx -1.285$, $x \approx 0.232$, and $x \approx 2.052$.

Step 4: Determine the sign of the expression in different intervals.

We now need to determine the sign of the expression $f(x) = x^4 - 6x^3 + 2x^2 + 9x - 8$ in different intervals to identify the regions where the inequality is satisfied.

Let's consider four intervals:

Interval 1: $x < -1.285$ Interval 2: $-1.285 < x < 0.232$ Interval 3: $0.232 < x < 2.052$ Interval 4: $x > 2.052$

Choose a test point within each interval and evaluate the expression $f(x)$ to determine its sign:

For $x = -2$, $f(-2) = 26 > 0$. (Interval 1 is positive) For $x = -1$, $f(-1) = 11 > 0$. (Interval 2 is positive) For $x = 1$, $f(1) = 3 > 0$. (Interval 3 is positive) For $x = 3$, $f(3) = 41 > 0$. (Interval 4 is positive)

Step 5: Identify the intervals that satisfy the original inequality.

Recall that our original inequality was:

$x^4 - 6x^3 + 2x^2 + 9x - 8 \geq 0$

Since the inequality includes "greater than or equal to," we need to find the intervals where the expression $f(x)$ is non-negative (positive or zero) because those are the regions that satisfy the inequality.

From our analysis in Step 4, we found that the expression $f(x)$ is positive in all four intervals. Therefore, the solution to the inequality is the entire real number line:

$x \in (-\infty, \infty)$

In summary, the solution to the inequality $(x^2-3x+1)(x^2-3x-3) \geq 5$ is $x \in (-\infty, \infty)$, which means that the inequality is satisfied for all real values of $x$.