Question

1) 2x² + 13x - 7 ≤ 0 2) ( x - 5 )² ( x² - 2x - 3) < 0 3) x² + 6x +9 4) x² + 6x +9 15 - 3x ≤ 0 _________ > 0 ________ <0 5) x + 2 ≥ 4x - 10 6) x² - 4x x² + 3x - 10 x² + 3x - 10 _____ _____ ______ ≤ 3 x - 2 x - 2 x - 2

Answer 1

Log5 (2x²+6x) ≤ log5 (2x²-3x)

(2x²+6x) ≤ (2x²-3x)
6х ≤ -3х
9х ≤ 0
х ≤ 0
образец

Answer 2

1) The inequality 2x² + 13x - 7 ≤ 0:

To solve this inequality, we can first find the critical points by setting the expression equal to zero:

2x² + 13x - 7 = 0

Next, factorize the quadratic equation or use the quadratic formula to find the solutions for x. Let's use the quadratic formula:

x = (-b ± √(b² - 4ac)) / (2a)

For our equation, a = 2, b = 13, and c = -7. Plugging in these values, we get:

x = (-13 ± √(13² - 4 * 2 * -7)) / (2 * 2) x = (-13 ± √(169 + 56)) / 4 x = (-13 ± √225) / 4 x = (-13 ± 15) / 4

So, the solutions for x are x = (-13 + 15) / 4 and x = (-13 - 15) / 4:

x = 2/4 = 1/2 x = -28/4 = -7

Now, we can plot the critical points on a number line and test the intervals between them:

-∞ (-7) (1/2) +∞

We can choose test points within each interval and substitute them into the original inequality to determine whether the inequality is satisfied or not. For example, we can choose x = 0:

2(0)² + 13(0) - 7 ≤ 0 -7 ≤ 0

Since -7 is less than or equal to 0, the inequality is satisfied for this interval.

Thus, the solution to the inequality 2x² + 13x - 7 ≤ 0 is:

x ≤ -7 or x ≥ 1/2

2) The inequality (x - 5)² + (x² - 2x - 3) < 0:

To solve this inequality, we can expand and simplify the left-hand side:

(x - 5)² + (x² - 2x - 3) < 0 (x² - 10x + 25) + (x² - 2x - 3) < 0 2x² - 12x + 22 < 0

Unfortunately, this inequality does not factorize easily. We can use the quadratic formula or complete the square to find the critical points. Let's use the quadratic formula:

x = (-b ± √(b² - 4ac)) / (2a)

For our equation, a = 2, b = -12, and c = 22. Plugging in these values, we get:

x = (-(-12) ± √((-12)² - 4 * 2 * 22)) / (2 * 2) x = (12 ± √(144 - 176)) / 4 x = (12 ± √(-32)) / 4

Since the discriminant (b² - 4ac) is negative, the quadratic equation has no real solutions. Therefore, there are no critical points and the inequality (x - 5)² + (x² - 2x - 3) < 0 is never true for any value of x.

Hence, the solution to the inequality (x - 5)² + (x² - 2x - 3) < 0 is:

No solution

3) The inequality x² + 6x + 9 < 0:

This inequality can be factored as a perfect square:

(x + 3)² < 0

A perfect square can never be negative or equal to zero. Therefore, there are no solutions to this inequality.

Hence, the solution to the inequality x² + 6x + 9 < 0 is:

No solution

4) The inequality x² + 6x + 9 + 15 - 3x ≤ 0:

First, let's simplify the left-hand side of the inequality:

x² + 6x + 9 + 15 - 3x ≤ 0 x² + 3x + 24 ≤ 0

This inequality does not factorize easily, so let's find the critical points using the quadratic formula:

x = (-b ± √(b² - 4ac)) / (2a)

For our equation, a = 1, b = 3, and c = 24. Plugging in these values, we get:

x = (-3 ± √(3² - 4 * 1 * 24)) / (2 * 1) x = (-3 ± √(9 - 96)) / 2 x = (-3 ± √(-87)) / 2

Since the discriminant (b² - 4ac) is negative, the quadratic equation has no real solutions. Therefore, there are no critical points and the inequality x² + 6x + 9 + 15 - 3x ≤ 0 is always true for any value of x.

So, the solution to the inequality x² + 6x + 9 + 15 - 3x ≤ 0 is:

All real numbers x satisfy the inequality

5) The inequality x + 2 ≥ 4x - 10:

Let's solve this inequality by bringing all the x terms to one side and the constant terms to the other side:

x + 2 ≥ 4x - 10 x - 4x ≥ -10 - 2 -3x ≥ -12

Dividing both sides of the inequality by -3, we need to flip the inequality sign since we are dividing by a negative number:

x ≤ (-12)/(-3) x ≤ 4

So, the solution to the inequality x + 2 ≥ 4x - 10 is:

x ≤ 4

6) The inequality x² - 4x + x² + 3x - 10 + x² + 3x - 10 ≤ 3 + x - 2 + x - 2 + x - 2:

Let's simplify both sides of the inequality:

x² - 4x + x² + 3x - 10 + x² + 3x - 10 ≤ 3 + x - 2 + x - 2 + x - 2 3x² + 2x - 20 ≤ 3

Now, bring all the terms to one side:

3x² + 2x - 20 - 3 ≤ 0 3x² + 2x - 23 ≤ 0

This quadratic inequality does not factorize easily, so let's find the critical points using the quadratic formula:

x = (-b ± √(b² - 4ac)) / (2a)

For our equation, a = 3, b = 2, and c = -23. Plugging in these values, we get:

x = (-2 ± √(2² - 4 * 3 * -23)) / (2 * 3) x = (-2 ± √(4 + 276)) / 6 x = (-2 ± √280) / 6

Now, we can simplify the expression under the square root:

√280 = √(2 * 2 * 2 * 5 * 7) = 2√70

Substituting this back into the equation, we get:

x = (-2 ± 2√70) / 6 x = (-1 ± √70) / 3

So, the critical points are x = (-1 + √70) / 3 and x = (-1 - √70) / 3.

Now, we can plot the critical points on a number line and test the intervals between them:

-∞ (-1 - √70)/3 (-1 + √70)/3 +∞

We can choose test points within each interval and substitute them into the original inequality to determine whether the inequality is satisfied or not. For example, we can choose x = 0:

3(0)² + 2(0) - 20 ≤ 3 -20 ≤ 3

Since -20 is not less than or equal to 3, the inequality is not satisfied for this interval.

Thus, the solution to the inequality x² - 4x + x² + 3x - 10 + x² + 3x - 10 ≤ 3 + x - 2 + x - 2 + x - 2 is:

No solution