Solve the inequation:    6 < 3x^2 -6 <  21

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giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll have to solve a system of 2 inequalities:

3x^2 -6 <  21

and

3x^2 -6 > 6

We'll solve the first inequality:

3x^2 -6 <  21

We'll add 6 both sides:

3x^2 < 27

We'll divide by 3:

x^2 < 9

We'll calculate the roots of x^2 - 9.

x^2 - 9 = 0

 x1 = -3

x2 = 3

The expression is negative over the interval (-3,3).

We'll solve the second inequality:

3x^2 -6 > 6

We'll add 6 both sides:

3x^2 > 12

We'll divide by 3:

x^2 > 4

We'll calculate the roots of x^2 - 4.

x^2 - 4 = 0

x1  = -2

x2 = 2

The expression is positive over the intervals (-inf.,-2)U(2,+inf.).

The common intervals of values that satisfy both inequalities are:

(-3 , -2)U(2 , 3)

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

6 < 3x^2-6 < 21.

We add 6 throught to each term: Then the inequality becomes:

0 < 3x^2< 21-6 = 15.

Divide by 6:

0 <x^2 < 5.

Take  square root:

0 <x < +sqrt5.........(1)

Or

-sqrt5< x < 0............(2)

Combining (1) and (2), we get:

-sqrt5 < x < +sqrt5.

william1941's profile pic

william1941 | College Teacher | (Level 3) Valedictorian

Posted on

We have to solve:    6 < 3x^2 -6 <  21

First let’s take 3x^2 -6 > 6

=> 3x^2 – 12 >0

=> 3 ( x^2 -4 ) >0

=> x^2 > 4

=> x >2 or x< -2

Second take 3x^2 -6 <  21

=> 3x^2 – 27 <0

=> 3(x^2 -9) <0

=> x^2 -9 <0

=> x^2 < 9

=> x <3 or x >-3

Now we have x>2 , x>-3 , x <3 and x < -2

So x can be between -3 and -2 or between 2 and 3 to meet all the criteria of the inequality.

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