# Which are the elements of the set A if they are solution of the inequality x^2 - 13x/3 - 10/3 > 0 ?

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x^2 - 13x/3 - 10/3 >0

to find the solution first we will solve the equation:

Multiply by 3:

==> 3x^2 - 13x - 10 > 0

Now factor:

==> (3x +2)(x-5) > 0

Tehn we have two cases:

case 1:

3x + 2 > 0 and x-5 > 0

x > -2/3 and x> 5

Then x belongs to (5, inf) ........(1)

case 2:

3x + 2 < 0 and x-5 <0

x < -2/3 and x < 5

Then x belongs to (-inf, -2/3) .........(2)

from (1) and (2) :

**x belongs to (-inf, -2/3) U (5, inf)**

To determine the elements of the set, we'll sove the inequality.

For this purpose, will solve first the equation:

x^2 - 13x/3 - 10/3 = 0

3x^2 - 13x - 10 = 0

After finding the roots of the equation, we could write the expression in a factored form as:

3(x-x1)(x-x2)>0

So, let's apply the quadratic formula to calculate the roots:

x1 = [13+sqrt(169-120)]/6

x1 = (13+sqrt49)/6

x1 = (13+7)/6

**x1 = 10/3**

x2 = (13-7)/6

x2 = 6/6

**x2 = 1**

The inequality will be written as:

3(x - 10/3)(x - 1) > 0

We'll divide by 3, both sides:

(x - 10/3)(x - 1) > 0

Now, we'll discuss the inequality:

- the product is positive if the factors are both positive:

x - 10/3>0

We'll add 10/3 both sides:

x > 10/3

and

x - 1 > 0

We'll add 1 both sides:

x > 1

So, x belongs to the interval (10/3 , +inf.)

- the product is positive if the factors are both negative:

x - 10/3 < 0

We'll add 10/3 both sides:

x < 10/3

x - 1 < 0

We'll add 1 both sides:

x < 1

So, x belongs to the interval ( -inf.,1)

Finally, the solution set of the inequality is the union of the sets above:

**( -inf.,1) U (10/3 , +inf.)**

**So, the set A is the union of intervals:**

**A = { ( -inf.,1) U (10/3 , +inf.)}**

x^2-13x/3 -10/3 > 0

To find the solution set:

Solution:

We first find the solution of the equation

x^2-13x/3 -10/3 = 0. Multiply by 3.

3x^2-13x-13 = 0.

(3x+2 )(x -5) = 0

Therefore 3x+2 = 0 or x-5 = 0. Or the roots are x= -2/3 or x= 5

So if x belongs to x <-2/3 then (3x+2)(x-5) > 0 as both factors are negative and the product is positive.

If x > 5, then both factors are positive and the product (3x+2)(x-5) > 0.

Therefore the solution set is x < -2/3 and x > 5.