Question

Система x^2+4x<1 x^2+4x>-1

Answer 1

$\left \{ {{x^2+4x-1}} \right.$

$\left \{ {{x^2+4x-10}} \right.$

решим оба квадратных уравнения:

1)x^2+4x-1<0;

D=16+4*1=20=4*5;

x1=(-4+2√5)/2=-2+√5;

x1=(-4-2√5)/2=-2-√5;

  +               -                  +

___-2-√5____-2+√5____

-2-√5<x<-2+√5;

2)x^2+4x+1>0;

D=16-4*1=12=4*3;

x1=(-4+2√3)/2=-2+√3;

x1=(-4-2√3)/2=-2-√3;

  +               -                  +

___-2-√3____-2+√3____

x<-2-√3;

x>-2+√3;

x ∈ (-2-√5;-2-√3) ∪ (-2+√3;-2+√5);

Answer 2

To solve the inequalities x^2 + 4x < 1 and x^2 + 4x > -1, we need to find the values of x that satisfy each inequality separately. Let's solve them step by step:

1. x^2 + 4x < 1:

To solve this inequality, we'll first move all terms to one side to get a quadratic inequality in standard form:

x^2 + 4x - 1 < 0

Now, let's find the critical points by setting the expression inside the inequality to zero and solving for x:

x^2 + 4x - 1 = 0

Using the quadratic formula, we get:

x = (-4 ± √(4^2 - 41(-1))) / 2*1 x = (-4 ± √(16 + 4)) / 2 x = (-4 ± √20) / 2 x = (-4 ± 2√5) / 2 x = -2 ± √5

So, the critical points are approximately x = -2 - √5 and x = -2 + √5.

Next, we'll test intervals to determine where the inequality is satisfied. We can use test points from each interval:

1. Test x = -3 (choose a value less than -2 - √5): x^2 + 4x = (-3)^2 + 4*(-3) = 9 - 12 = -3 (which is < 1)

2. Test x = 0 (choose a value between -2 - √5 and -2 + √5): x^2 + 4x = 0^2 + 4*0 = 0 (which is < 1)

3. Test x = 1 (choose a value greater than -2 + √5): x^2 + 4x = 1^2 + 4*1 = 1 + 4 = 5 (which is NOT < 1)

Now, we can determine the solution based on the tests:

-∞ < x < -2 - √5 or -2 + √5 < x < ∞

2. x^2 + 4x > -1:

To solve this inequality, we'll first move all terms to one side:

x^2 + 4x + 1 > 0

Now, let's find the critical points by setting the expression inside the inequality to zero and solving for x:

x^2 + 4x + 1 = 0

Using the quadratic formula, we get:

x = (-4 ± √(4^2 - 411)) / 2*1 x = (-4 ± √(16 - 4)) / 2 x = (-4 ± √12) / 2 x = (-4 ± 2√3) / 2 x = -2 ± √3

So, the critical points are approximately x = -2 - √3 and x = -2 + √3.

Next, we'll test intervals to determine where the inequality is satisfied:

1. Test x = -3 (choose a value less than -2 - √3): x^2 + 4x = (-3)^2 + 4*(-3) + 1 = 9 - 12 + 1 = -2 (which is > -1)

2. Test x = 0 (choose a value between -2 - √3 and -2 + √3): x^2 + 4x = 0^2 + 4*0 + 1 = 1 (which is > -1)

3. Test x = 1 (choose a value greater than -2 + √3): x^2 + 4x = 1^2 + 4*1 + 1 = 6 (which is > -1)

Now, we can determine the solution based on the tests:

-∞ < x < -2 - √3 or -2 + √3 < x < ∞

Combining both solutions, we get the final answer:

-∞ < x < -2 - √5 or -2 + √5 < x < ∞