Question

36>X^2 X^2-8x<0 2x^2+5x-3>0

Answer 1

Ответ:

Объяснение:

36>x²

x²-36<0

x²-6²<0

(x+6)(x-6)<0

-∞__+__-6__-__6__+__+∞

x∈(-6;6).

x²-8x<0

x*(x-8)<0

-∞__+__0__-__8__+__+∞

x∈(0;8).

2x²+5x-3>0

2x²+5x-3=0   D=49    √D=7

x₁=-3        x₂=0,5    ⇒

(x+3)(x-0,5)>0

-∞__+__-3__-__0,5__+__+∞

x∈(-∞;-3)U(0,5;+∞).

Answer 2

To solve the inequalities:

1. 36 > X^2

We subtract 36 from both sides:

X^2 - 36 > 0

Now we factor the equation:

(X - 6)(X + 6) > 0

The critical points are X = -6 and X = 6. We can create a sign chart to analyze the inequality:

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(-∞)----(-6)----(6)----(+∞)
      -     +     -     

From the sign chart, we can see that the inequality is satisfied when X is in the interval (-∞, -6) U (6, +∞).

2. X^2 - 8X < 0

We factor the equation:

X(X - 8) < 0

The critical points are X = 0 and X = 8. We create another sign chart to analyze the inequality:

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(-∞)----(0)----(8)----(+∞)
      -     +     -     

From the sign chart, we can see that the inequality is satisfied when X is in the interval (0, 8).

3. 2X^2 + 5X - 3 > 0

We can factor the quadratic equation:

(2X - 1)(X + 3) > 0

The critical points are X = 1/2 and X = -3. We create the final sign chart to analyze the inequality:

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(-∞)----(-3)----(1/2)----(+∞)
      -     +      -     

From the sign chart, we can see that the inequality is satisfied when X is in the intervals (-∞, -3) U (1/2, +∞).

In summary, the solutions to the given inequalities are:

1. X is in the interval (-∞, -6) U (6, +∞).
2. X is in the interval (0, 8).
3. X is in the intervals (-∞, -3) U (1/2, +∞).