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andriod09

  • 2 years ago

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  1. andriod09
    • 2 years ago
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    \[\sqrt{2y-3}+3=y\]

  2. andriod09
    • 2 years ago
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    @cshalvey

  3. andriod09
    • 2 years ago
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    @.UserNotFound. @jazy @cshalvey

  4. andriod09
    • 2 years ago
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    @KingGeorge

  5. KingGeorge
    • 2 years ago
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    In the future, please refrain from tagging a bunch of users. Anyways, first move the 3 over.\[\sqrt{2y-3}+3=y\]\[\sqrt{2y-3}=y-3\]Now square both sides. Can you tell me what youget?

  6. andriod09
    • 2 years ago
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    sorry, they're all the people that i have fanned for helping me.

  7. KingGeorge
    • 2 years ago
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    No worries, just something to remember for the future. Anyway, can you tell me what you get after squaring both sides?

  8. andriod09
    • 2 years ago
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    \[\left( \sqrt{2y-3} \right)^{2}=\left( y+3 \right)^{2}\] \[\left( 2y-3 \right)=\left(y ^{2}+9 \right)\]

  9. KingGeorge
    • 2 years ago
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    The left side looks good. The right is not correct. Remember that you need to FOIL. Can you try again?

  10. andriod09
    • 2 years ago
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    whats a "FOIL"?

  11. KingGeorge
    • 2 years ago
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    It's an acronym for First Outer Inner Last It means you first multiply the first two numbers together, then the outer numbers together, then the inner, and then the last numbers. Once you've done that, you add them all. For example: \[(y+3)(y+2)\] First: \((\color{red}y+3)(\color{red}y+2)\implies y\cdot y=y^2\) Outer: \((\color{red}y+3)(y+\color{red}2)\implies y\cdot 2=2y\) Inner: \((y+\color{red}3)(\color{red}y+2)\implies 3\cdot y=3y\) Last: \((y+\color{red}3)(y+\color{red}2)\implies 3\cdot2=6\) So \((y+3)(y+2)=y^2+2y+3y+6=y^2+5y+6\)

  12. KingGeorge
    • 2 years ago
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    Now can you try the same thing with \((y+3)^2=(y+3)(y+3)\)?

  13. KingGeorge
    • 2 years ago
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    Oh, and it should be \((y-3)^2=(y-3)(y-3)\). Oops :(

  14. andriod09
    • 2 years ago
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    \[\left( y+3 \right)\left( y+3 \right)\] \[y^{2}+3y+3y+9?\]

  15. KingGeorge
    • 2 years ago
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    Right, and if you had \((y-3)(y-3)\), it would become something similar. You get \[y^2-3y-3y+9=y^2-6y+9\] This means you have \[2y-3=y^2-6y+9\]Get 0 on one side, and you get \[y^2-8y+12=0\]Now you just need to factor this. Can you manage that yourself?

  16. andriod09
    • 2 years ago
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    yes i can. Thank you soo much king. for every thing. ima fan you

  17. KingGeorge
    • 2 years ago
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    You're welcome. Feel free to post what you get as solutions and I can check them if you want.

  18. andriod09
    • 2 years ago
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    kk will do. Thanks

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