Question

(x-3)√(14+5x-x²) < 0 інтервали

Answer 1

Ответ:  [-2;3)

Объяснение:

Answer 2

Вот))))))))))))))))))))

Answer 3

To solve the inequality $(x-3)\sqrt{14+5x-x^2} < 0$, we need to find the intervals where the expression on the left-hand side is negative.

Step 1: Determine the critical points To find the critical points, set the expression inside the square root equal to zero and solve for $x$:

$14 + 5x - x^2 = 0$

This is a quadratic equation, so let's rearrange it:

$x^2 - 5x - 14 = 0$

Now, we can factor this equation:

$(x - 7)(x + 2) = 0$

Setting each factor to zero gives us the critical points:

$x - 7 = 0$ $\implies x = 7$

$x + 2 = 0$ $\implies x = -2$

Step 2: Create test intervals Now, we need to choose test intervals to determine the sign of the expression $(x-3)\sqrt{14+5x-x^2}$. The test intervals are determined by the critical points found in Step 1.

Test Interval 1: $x < -2$

Test Interval 2: $-2 < x < 7$

Test Interval 3: $x > 7$

Step 3: Evaluate the expression in each test interval Now, we evaluate the expression $(x-3)\sqrt{14+5x-x^2}$ in each test interval:

For Test Interval 1 ($x < -2$): Choose $x = -3$ (a value less than -2): $(-3 - 3)\sqrt{14+5(-3)-(-3)^2} = (-6)\sqrt{14-15} = (-6)\sqrt{-1} = -6i$ (where $i$ is the imaginary unit)

For Test Interval 2 ($-2 < x < 7$): Choose $x = 0$ (a value between -2 and 7): $(0 - 3)\sqrt{14+5(0)-0^2} = (-3)\sqrt{14} < 0$ (since the square root of a positive number is positive)

For Test Interval 3 ($x > 7$): Choose $x = 8$ (a value greater than 7): $(8 - 3)\sqrt{14+5(8)-8^2} = (5)\sqrt{14+40-64} = (5)\sqrt{-10} = 5i$ (where $i$ is the imaginary unit)

Step 4: Determine the solution Since we are looking for the interval where $(x-3)\sqrt{14+5x-x^2} < 0$, the solution is the interval where the expression is negative.

From the evaluations above, we can see that the expression is negative only in Test Interval 1, which corresponds to the values of $x < -2$.

So, the solution to the inequality is:

$x < -2$