Question

Solving inequalities... help!
(x-10)^3 > 53

A. x < -13.756 or x > 13.756
B. x < -7.28 or x > 7.28
C. x > 13.756
D. x > 15.418

Answer 1

Alright so let's find the critical points of the inequality
\[\ x^3−30x^2+300x−1000=53\]

Answer 2

So subtract 53 from both sides
\[\ x^3−30x^2+300x−1000−\color{red}{53}=53−\color{red}{53}\]
\[\ x^3−30x^2+300x−1053=0\]

Answer 3

I believe you know what cubic formula is use it \(x=13.756286\)
Also don't forget to check intervals in between critical points.

Answer 4

and your final answer should be \(x>13.756286\) since that would work in original inequality.

Answer 5

@hannahlb99 when you get online tell me if you understand this.

Answer 6

okay! thank you so much

Answer 7

cube of a negative number is negative
and cube of a positive number is positive.
as right hand side is positive so x-10>0
x>10
\[x-10>53^{\frac{ 1 }{ 3 }}\]
\[x-10>3.75628575\]
x>10+3.75628575
?