X^2+7x<0 2x^2-x-6 =0

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вопрос

To solve the inequality x^2 + 7x < 0, we need to find the values of x that make the expression less than 0.

First, let's factor the expression x^2 + 7x:

x(x+7) < 0

Now we need to find the critical points where x(x+7) = 0:

x = 0 and x = -7

These points divide the number line into three intervals:

We can now test each interval by plugging in any number from that interval into x(x+7) to see if it is less than 0.

For x < -7, let's choose x = -8:
-8(-8+7) = -8(-1) = 8, which is not less than 0.

For -7 < x < 0, let's choose x = -1:
-1(-1+7) = -1(6) = -6, which is less than 0.

For x > 0, let's choose x = 1:
1(1+7) = 1(8) = 8, which is not less than 0.

Therefore, the solution to the inequality x^2 + 7x < 0 is -7 < x < 0.

Next, let's solve the quadratic equation 2x^2 - x - 6 = 0:

We can factor this quadratic equation as follows:
2x^2 - x - 6 = 0
(2x + 3)(x - 2) = 0

Setting each factor equal to 0:
2x + 3 = 0 or x - 2 = 0

Solving these equations gives us:
2x = -3 or x = 2
x = -3/2 x = 2

Therefore, the solutions to the quadratic equation 2x^2 - x - 6 = 0 are x = -3/2 and x = 2.