Castiel
  • Castiel
Solve inequality |(x+3)^2|<1
Mathematics
schrodinger
  • schrodinger
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Castiel
  • Castiel
The exponent is confusing me
IrishBoy123
  • IrishBoy123
get rid of the || first, its redundant then square root?!
anonymous
  • anonymous
Let's solve your inequality step-by-step. (x+3)2<1 x2+6x+9<1 Let's find the critical points of the inequality. x2+6x+9=1 x2+6x+9−1=1−1(Subtract 1 from both sides) x2+6x+8=0 (x+2)(x+4)=0(Factor left side of equation) x+2=0 or x+4=0(Set factors equal to 0) x=−2 or x=−4 Check intervals in between critical points. (Test values in the intervals to see if they work.) x<−4(Doesn't work in original inequality) −4−2(Doesn't work in original inequality) Answer: −4

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Castiel
  • Castiel
so basically because of the exponent it's the same as |(x+3)|<1
anonymous
  • anonymous
yeahh
IrishBoy123
  • IrishBoy123
no the square will be +ve so you don't need the | |, it's redundant so get rid of it first
anonymous
  • anonymous
yeahh thats what throws it off my fault
IrishBoy123
  • IrishBoy123
Casteil try doing it yourself having got rid of the ||, we can addresss the exponent directly and in far far fewer lines!!!
IrishBoy123
  • IrishBoy123
without the ||'s, what does the equation now say?
Castiel
  • Castiel
but square root of (x+3)^3 is again |x+3|
Castiel
  • Castiel
^2 not ^3
anonymous
  • anonymous
Right
IrishBoy123
  • IrishBoy123
if you know that \((x+3)^2 < 1\) then you know that \( |x+3|<1\), yes?
Castiel
  • Castiel
and then from that -1
anonymous
  • anonymous
yes
IrishBoy123
  • IrishBoy123
well done castiel!

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