Solve the inequality: SQRT (3x - 2)2 < 5

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

We have to solve sqrt ( 3x - 2)^2 < 5

=> (3x - 2)^2 < 25

Now (3x - 2)^2 can be less than 25 if 3x - 2 is less than sqrt 5 or if 3x - 2 is greater than - sqrt 5

This gives two inequalities

3x - 2 < 5

=> 3x < 7

=> x < 7/3

and 3x - 2 > -5

=> 3x > -3

=> x > -3/3

=> x > -1

So we have -1 < x < 7/3

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hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

Given the equation:

sqrt(3x-2)^2 < 5

Now we will rewrite using the absolute value form.

==> l 3x-2 l < 5

Now we will rewrite :

-5 < 3x-2 < 5

Now we will add 2 to both sides.

==> -3 < 3x < 7

Now we will divide by 3.

==> -1 < x < 7/3

Then the answer is: x= ( -1, 7/3).

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wallahi77 | Student, Undergraduate | eNotes Newbie

Posted on

SQRT(3x-2)*2<5

SQRT(3x-2)<5/2

3x-2 < (2.5)^2

3x < 6.25+2

x < 8.25/3

x < 2.75

 

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