Question

A)2x2 – 13x + 6 < 0;​

Answer 1

Ответ:

Пошаговое объяснение:

Answer 2

To solve the inequality $2x^2 - 13x + 6 < 0$, you can follow these steps:

1. First, find the roots of the quadratic equation $2x^2 - 13x + 6 = 0$. You can do this by factoring or by using the quadratic formula:

   $2x^2 - 13x + 6 = 0$

   Factoring: $(2x - 1)(x - 6) = 0$

   Setting each factor equal to zero: $2x - 1 = 0$ or $x - 6 = 0$

   Solve for x: $2x = 1$ or $x = 6$

   So, the roots of the equation are $x = 1/2$ and $x = 6$.

2. Now, you have three critical points on the number line: $x = 1/2$, $x = 6$, and any other point you choose (since quadratic inequalities involve parabolas and open intervals, you can choose any point to test).

3. Create a number line and plot these critical points on it:

   luaCopy code
   |-----------------------|-------------------|
   0                     1/2                 6
   

4. Now, you need to test the regions created by these critical points to determine when $2x^2 - 13x + 6$ is less than zero.

   • Choose a test point in the region to the left of $1/2$, let's say $x = 0$.

   • Plug this value into the original inequality:

     $2(0)^2 - 13(0) + 6 < 0$

     $6 < 0$

   Since $6$ is not less than zero, this region is not part of the solution.

   • Choose a test point in the region between $1/2$ and $6$, let's say $x = 3$.

   • Plug this value into the original inequality:

     $2(3)^2 - 13(3) + 6 < 0$

     $18 - 39 + 6 < 0$

     $-15 + 6 < 0$

     $-9 < 0$

   Since $-9$ is less than zero, this region is part of the solution.

   • Choose a test point in the region to the right of $6$, let's say $x = 7$.

   • Plug this value into the original inequality:

     $2(7)^2 - 13(7) + 6 < 0$

     $98 - 91 + 6 < 0$

     $7 + 6 < 0$

     $13 < 0$

   Since $13$ is not less than zero, this region is not part of the solution.

5. Based on the tests, the solution to the inequality $2x^2 - 13x + 6 < 0$ is:

   $\frac{1}{2} < x < 6$

So, the values of $x$ that satisfy the inequality are $x$ in the open interval $\left(\frac{1}{2}, 6\right)$.