Question

Решите неравенство: 5x^2-11x+2≥0

Answer 1

$5 {x}^{2} - 11x + 2 \geqslant 0 \\ d = {b}^{2} - 4ac = 121 - 4 \times 5 \times 2 = 81 \\ x1 = \frac{11 + 9}{2 \times 5} = \frac{20}{10} = 2 \\ x2 = \frac{11 - 9}{2 \times 5} = \frac{2}{10} = \frac{1}{5}$
Неравенство примет вид:
5*(x - 2)*(x - 1/5) >= 0
Отметим точки на координатной прямой (далее во вложении)

Answer 2

5*12-11x+2≥0

60-11x+2≥0

62-11x≥0

-11x≥-62(:-11)

x≥62/11

Answer 3

To solve the inequality $5x^2 - 11x + 2 \geq 0$, we need to find the values of $x$ that satisfy this inequality.

Step 1: Find the roots of the quadratic equation $5x^2 - 11x + 2 = 0$. To do this, we can use the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $ax^2 + bx + c = 0$ is the standard form of a quadratic equation.

For our equation $5x^2 - 11x + 2 = 0$, we have $a = 5$, $b = -11$, and $c = 2$. Using the quadratic formula:

$x = \frac{11 \pm \sqrt{(-11)^2 - 4 \cdot 5 \cdot 2}}{2 \cdot 5} = \frac{11 \pm \sqrt{121 - 40}}{10} = \frac{11 \pm \sqrt{81}}{10} = \frac{11 \pm 9}{10}$

So the roots are $x = \frac{20}{10} = 2$ and $x = \frac{2}{10} = \frac{1}{5}$.

Step 2: Determine the sign of $5x^2 - 11x + 2$ in the intervals created by the roots.

To do this, we can create a sign chart:

$$\begin{array}{|c|c|c|c|c|} \hline \text{Interval} & ( -\infty, \frac{1}{5}) & (\frac{1}{5}, 2) & (2, \infty) \\ \hline \text{Test Point} & 0 & 1 & 3 \\ \hline \text{Sign of } 5x^2 - 11x + 2 & + & + & + \\ \hline \end{array}$$

Step 3: Analyze the inequality based on the sign chart.

We want to find the values of $x$ for which $5x^2 - 11x + 2 \geq 0$, i.e., the regions where the expression is greater than or equal to zero. Looking at the sign chart, we see that the expression is always positive in all three intervals.

Step 4: Write the final solution.

Since $5x^2 - 11x + 2$ is always positive ($\geq 0$) for all real values of $x$, the solution to the inequality is the set of all real numbers:

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

In interval notation, the solution is $(- \infty, \infty)$.