Question

3x^ - 11x + 6 < 0 решите неравенство

Answer 1

$3x^{2} - 11x + 6 < 0 \\ d = 121 - 4 \times 3 \times 6 = 49 \\ \sqrt{d} = 7 \\ x1 = \frac{ 11 + 7}{2 \times 3} = 3 \\ x2 = \frac{2}{3} \\ (( \frac{2}{3} )(3))$

Answer 2

To solve the inequality $3x^2 - 11x + 6 < 0$, we can use the method of factoring. First, we factor the quadratic expression on the left-hand side and then find the intervals where the expression is less than zero.

Step 1: Factor the quadratic expression. The given quadratic expression is $3x^2 - 11x + 6$. We need to find two binomials whose product equals this expression. The factored form will be in the form $(ax + b)(cx + d)$.

$3x^2 - 11x + 6 = (3x - 2)(x - 3)$

Step 2: Set each factor to zero and solve for $x$. Set $3x - 2$ and $x - 3$ to zero and solve for $x$:

$3x - 2 = 0$ => $x = \frac{2}{3}$

$x - 3 = 0$ => $x = 3$

Step 3: Create a sign chart to find the intervals where the expression is less than zero.

We have three critical points: $x = \frac{2}{3}$, $x = 3$, and we can choose any other value in the real number line as a test point. Let's choose $x = 0$ as the test point.

Now, construct a sign chart:

$$\begin{array}{|c|c|c|c|c|} \hline \text{Interval} & ( -\infty, \frac{2}{3} ) & (\frac{2}{3}, 3) & (3, \infty) \\ \hline \text{Test point} & x = 0 & x = 1 & x = 4 \\ \hline \text{Sign of }(3x^2 - 11x + 6) & + & - & + \\ \hline \end{array}$$

Step 4: Interpret the results. From the sign chart, we can see that the expression $3x^2 - 11x + 6$ is less than zero ($< 0$) in the interval $\left(\frac{2}{3}, 3\right)$.

Step 5: Write the solution. The solution to the inequality is $x \in \left(\frac{2}{3}, 3\right)$.

In interval notation, the solution is $x \in \left(\frac{2}{3}, 3\right)$. This means any value of $x$ in the open interval between $\frac{2}{3}$ and $3$ will satisfy the inequality.