first year calculus HELPP!! :(

[luvablearab]

Find all real values of x such that

1?(2x-3)² > 1/4

can someone please help me, i dont know where to start..

 

[Deleted member 4993]

Find all real values of x such that

1?(2x-3)² > 1/4

can someone please help me, i dont know where to start..

Please check your post - read it carefully - and confirm that it is correct (in case it is incorrect - edit and correct it)

 

[luvablearab]

oh opps

it is incorrect.

this is the correct question:

Find all real values of x such that

1/?(2x-3)² > 1/4

 

[mmm4444bot]

Write the reciprocal of both sides, and change the direction of the inequality symbol.

EG:

1/2 > 1/4

2/1 < 4/1

Also, remember that the square root of an algebraic expression like (2x - 3)^2 is |2x - 3|, so you need to consider both the principal square root and its opposite.

Does this help ?

 

[BigGlenntheHeavy]

\(\displaystyle \frac{1}{\sqrt{(2x-3)^2}} \ > \ \frac{1}{4}, \ \implies \ |2x-3| \ < \ 4.\)

\(\displaystyle -4 \ < \ 2x-3 \ < \ 4, \ \frac{-1}{2} \ < \ x \ < \ \frac{7}{2}, \ x \ \ne \ \frac{3}{2}.\)

 

[Deleted member 4993]

In addition

\(\displaystyle (2x-3) \ \ne \ 0\)

 

[mmm4444bot]

\(\displaystyle (2x-3) \ \ne \ 0\)

Good catch! The solution is a union of two sets.