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anonymous

  • one year ago

please help !!!!! Choose any four (4) of the five radical equations below. Solve each equation, showing all steps leading to your answer. Then check your answer to determine if any of the solutions are extraneous i just need you guys to help me in one question 4. 2√4 - 2x = 2 - x please help !!! how do we solve it ?

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  1. random231
    • one year ago
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    |dw:1441477010872:dw| now square both sides and solve the eqn

  2. Jhannybean
    • one year ago
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    \[2\sqrt{4}-2x=2-x\] Rearrange.\[2\sqrt{4}-2=2x-x\]\[\boxed{x=2\sqrt{4}-2}\]

  3. IrishBoy123
    • one year ago
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    \[\sqrt{4}\] is this for real?!

  4. anonymous
    • one year ago
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    No, I think it's four real xD im sorry i had to

  5. Jhannybean
    • one year ago
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    You didnt simplify my answer? :(

  6. IrishBoy123
    • one year ago
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    it's "four real" !!! lol!!

  7. anonymous
    • one year ago
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    jhannybean, the 2x is also the under the radical. so can we really send it to the other side ?

  8. Jhannybean
    • one year ago
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    so your function is \(2\sqrt{4-2x}= 2-x\)?

  9. anonymous
    • one year ago
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    yeah

  10. Jhannybean
    • one year ago
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    Okay. Then we would have to begin by squaring both sides. OUr goal is to isolate the x's to one side.

  11. Jhannybean
    • one year ago
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    \[(2\sqrt{4-2x})^2 = (2-x)^2\]\[4(4-2x) = 4-4x+x^2\]\[16-8x=4-4x+x^2\]\[x^2-4x+8x=16-4\]\[x^2+4x-12=0\]Can you factor this now?

  12. anonymous
    • one year ago
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    um is it (x + 2)(x - 6) = 0 ?

  13. Jhannybean
    • one year ago
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    \[(x+2)(x-6) = x^2 -4x-12 \ne x^2+4x-12\]

  14. anonymous
    • one year ago
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    oh i think i got it is it (x - 2)( x + 6 ) ?

  15. Jhannybean
    • one year ago
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    Yep

  16. anonymous
    • one year ago
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    then we simplify like this : x - 2 = 0 x = 2 and x + 6 = 0 x = - 6 so the answer is x = 2, -6

  17. Jhannybean
    • one year ago
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    Mmhmm

  18. anonymous
    • one year ago
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    omg thank you so much :)

  19. Jhannybean
    • one year ago
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    No problem :)

  20. anonymous
    • one year ago
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    can i ask u one more question ?

  21. Jhannybean
    • one year ago
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    Sure

  22. anonymous
    • one year ago
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    it is similar, \[\sqrt{5x - x } = 0\]

  23. anonymous
    • one year ago
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    how do we solve this ?

  24. anonymous
    • one year ago
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    Square both sides first, right?

  25. Jhannybean
    • one year ago
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    Again, we want to isolate x so we need to square both sides. Same method.

  26. anonymous
    • one year ago
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    wouldnt it just be like 4x = 0 at the end? xD or just plain out 0?

  27. Jhannybean
    • one year ago
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    You are correct.

  28. anonymous
    • one year ago
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    Oooooooo, look at me im doin stuff xD

  29. anonymous
    • one year ago
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    oh sorry , made a mistake again i meant \[\sqrt{5x} - x = 0 \]

  30. anonymous
    • one year ago
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    Wouldnt that be the same outcome?

  31. anonymous
    • one year ago
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    Er... nevermind

  32. Jhannybean
    • one year ago
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    \[\sqrt{5x}-x=0\]\[\sqrt{5x}=x\]\[(\sqrt{5x})^2=x^2\]\[5x=x^2\]\[x^2-5x=0\]\[x(x-5)=0\]\[x=0\]\[x-5=0 \iff x=5\]

  33. Jhannybean
    • one year ago
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    @BweadedChicken no, not the same outccome.

  34. anonymous
    • one year ago
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    oh that was it ? i thought we need to complete the square and all any way , thank you you're the best :)

  35. Jhannybean
    • one year ago
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    Yeah, completing the square comes along with complex quadratic functions that need to be simplified. You'll know when you see them.

  36. anonymous
    • one year ago
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    oh ok :)

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