Question

2≤x²+x<6 розвязати ............

Answer 1

$2\leq x^2+x$

Answer 2

x^2+x≥2

x^2+x-2≥0

D=1+8=9

x1=(-1+3)/2*1=2/2=1

x2=(-1-3)/2*1=-4/2=-2

(-oo;-2]U[1;+oo)

x^2+x<6

x^2+x-6<0

D=1+24=25

x1=(-1+5)/2*1=4/2=2

x2=(-1-5)/2*1=-6/2=-3

(-3;2)

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______-3_____________-2__________1_________2_________

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(-3;-2]U[1;2)

Answer 3

x^2+x≥2

x^2+x-2≥0

D=1+8=9

x1=(-1+3)/2*1=2/2=1

x2=(-1-3)/2*1=-4/2=-2

(-oo;-2]U[1;+oo)

x^2+x<6

x^2+x-6<0

D=1+24=25

x1=(-1+5)/2*1=4/2=2

x2=(-1-5)/2*1=-6/2=-3

(-3;2)

///////////////////////////////////////////////////////////////////

______-3_____________-2__________1_________2_________

\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

(-3;-2]U[1;2)

Answer 4

To solve the inequality 2 ≤ x² + x < 6, we need to find the values of x that satisfy the inequality. Let's break it down into two separate inequalities:

1. x² + x ≥ 2
2. x² + x < 6

Step 1: Solve x² + x ≥ 2

To solve the first inequality, x² + x ≥ 2, we need to find the values of x that make the expression greater than or equal to 2. To do that, we'll move all terms to one side of the inequality:

x² + x - 2 ≥ 0

Now, we can factor the quadratic expression:

(x + 2)(x - 1) ≥ 0

To find the solution, we need to consider three intervals: x < -2, -2 ≤ x ≤ 1, and x > 1. We'll pick test points from each interval and check if they satisfy the inequality:

Test point x = -3 (x < -2): (-3 + 2)(-3 - 1) ≥ 0 (-1)(-4) ≥ 0 4 ≥ 0 (True)

Test point x = 0 (-2 ≤ x ≤ 1): (0 + 2)(0 - 1) ≥ 0 (2)(-1) ≥ 0 -2 ≥ 0 (False)

Test point x = 2 (x > 1): (2 + 2)(2 - 1) ≥ 0 (4)(1) ≥ 0 4 ≥ 0 (True)

The solutions for x² + x ≥ 2 are x ≤ -2 and x > 1.

Step 2: Solve x² + x < 6

To solve the second inequality, x² + x < 6, we need to find the values of x that make the expression less than 6. To do that, we'll move all terms to one side of the inequality:

x² + x - 6 < 0

Now, we can factor the quadratic expression:

(x + 3)(x - 2) < 0

To find the solution, we again need to consider three intervals: x < -3, -3 ≤ x ≤ 2, and x > 2. We'll pick test points from each interval and check if they satisfy the inequality:

Test point x = -4 (x < -3): (-4 + 3)(-4 - 2) < 0 (-1)(-6) < 0 6 < 0 (False)

Test point x = 0 (-3 ≤ x ≤ 2): (0 + 3)(0 - 2) < 0 (3)(-2) < 0 -6 < 0 (True)

Test point x = 3 (x > 2): (3 + 3)(3 - 2) < 0 (6)(1) < 0 6 < 0 (False)

The solutions for x² + x < 6 are -3 ≤ x ≤ 2.

Combining the results from both inequalities:

1. x ≤ -2 or x > 1
2. -3 ≤ x ≤ 2

The final solution for the original inequality 2 ≤ x² + x < 6 is:

-3 ≤ x ≤ 2, and x > 1.