Question

2х^2-7х+6 меньше или равно 0

Answer 1

2х^2-7х+6=0; D=(-7)^2-4*2*6=1; x1=(7+1)/4=2; x2=(7-1)/4=1,5; (x-1,5)*(x-2)<=0; [1,5; 2]

Answer 2

To find the values of x for which the inequality $2x^2 - 7x + 6 \leq 0$ is true, we can follow these steps:

Step 1: Find the roots of the quadratic equation $2x^2 - 7x + 6 = 0$. Step 2: Plot the graph of the quadratic equation. Step 3: Determine the regions where the graph is below or equal to zero.

Step 1: Find the roots of the quadratic equation $2x^2 - 7x + 6 = 0$: To find the roots, 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 $2x^2 - 7x + 6 = 0$, we have $a = 2$, $b = -7$, and $c = 6$.

Using the quadratic formula:

$x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4 \cdot 2 \cdot 6}}{2 \cdot 2}$

$x = \frac{7 \pm \sqrt{49 - 48}}{4}$

$x = \frac{7 \pm \sqrt{1}}{4}$

$x = \frac{7 \pm 1}{4}$

So the roots of the equation are $x = \frac{8}{4} = 2$ and $x = \frac{6}{4} = 1.5$.

Step 2: Plot the graph of the quadratic equation: The graph of the quadratic equation $2x^2 - 7x + 6$ is a parabola that opens upwards because the coefficient of $x^2$ is positive (2 in this case). The roots we found in Step 1 ($x = 2$ and $x = 1.5$) are the x-intercepts of the parabola.

Step 3: Determine the regions where the graph is below or equal to zero: To find the regions where the graph is less than or equal to zero, we need to identify the intervals on the x-axis where the graph is below or touches the x-axis.

Since the parabola opens upwards, the graph will be below the x-axis outside the interval between the roots $x = 2$ and $x = 1.5$. In other words, the inequality $2x^2 - 7x + 6 \leq 0$ is true for $x$ values in the interval $(-\infty, 1.5] \cup [2, \infty)$.

So, the solution to the inequality is $x \in (-\infty, 1.5] \cup [2, \infty)$.