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Castiel

  • one year ago

Solve inequality |(x+3)^2|<1

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  1. Castiel
    • one year ago
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    The exponent is confusing me

  2. IrishBoy123
    • one year ago
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    get rid of the || first, its redundant then square root?!

  3. anonymous
    • one year ago
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    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<x<−2(Works in original inequality) x>−2(Doesn't work in original inequality) Answer: −4<x<−2

  4. Castiel
    • one year ago
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    so basically because of the exponent it's the same as |(x+3)|<1

  5. anonymous
    • one year ago
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    yeahh

  6. IrishBoy123
    • one year ago
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    no the square will be +ve so you don't need the | |, it's redundant so get rid of it first

  7. anonymous
    • one year ago
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    yeahh thats what throws it off my fault

  8. IrishBoy123
    • one year ago
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    Casteil try doing it yourself having got rid of the ||, we can addresss the exponent directly and in far far fewer lines!!!

  9. IrishBoy123
    • one year ago
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    without the ||'s, what does the equation now say?

  10. Castiel
    • one year ago
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    but square root of (x+3)^3 is again |x+3|

  11. Castiel
    • one year ago
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    ^2 not ^3

  12. anonymous
    • one year ago
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    Right

  13. IrishBoy123
    • one year ago
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    if you know that \((x+3)^2 < 1\) then you know that \( |x+3|<1\), yes?

  14. Castiel
    • one year ago
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    and then from that -1<x+3<1 and we get -4<x<-2

  15. anonymous
    • one year ago
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    yes

  16. IrishBoy123
    • one year ago
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    well done castiel!

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