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need_help! Group Title

Solve (4x^2-3)^2=64

  • 2 years ago
  • 2 years ago

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  1. Retract Group Title
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    The square root and 2th power exclude each other as they are both doing the opposite. You therefore get: \[4x^2 - 3 = 64\]

    • 2 years ago
  2. shaun786 Group Title
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    4x^2 = 67 x^2 = 67/4 x = 67/4 , -67/4

    • 2 years ago
  3. need_help! Group Title
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    Thanks =)

    • 2 years ago
  4. need_help! Group Title
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    But 64 does not become 8?

    • 2 years ago
  5. Kreshnik Group Title
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    shaun has it right untill he gets rid of power it should be: \[\LARGE x^2=\frac{67}{4}\] then... \[\LARGE x=\pm \sqrt{\frac{67}{4}}\] but first of all... just want to ask you. is this the given: \[\LARGE \sqrt{(4x^2-3)^2}=64 \]

    • 2 years ago
  6. shaun786 Group Title
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    oh yeah forgot abt the squarerott. Thanks kreshnik

    • 2 years ago
  7. Kreshnik Group Title
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    :)

    • 2 years ago
  8. need_help! Group Title
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    Yes! That's the given.

    • 2 years ago
  9. need_help! Group Title
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    Actually I don't know whether the square root is for everything or only for (4x^2-3)^2

    • 2 years ago
  10. need_help! Group Title
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    But it does matter, right?

    • 2 years ago
  11. need_help! Group Title
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    I mean if it was for everything then 64 will become 8?

    • 2 years ago
  12. lalaly Group Title
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    What do u mean for everything?

    • 2 years ago
  13. need_help! Group Title
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    for the whole equation

    • 2 years ago
  14. lalaly Group Title
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    the whole sqrt is squared? \[(\sqrt{4x^2-3})^2=64\]

    • 2 years ago
  15. need_help! Group Title
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    Okay, sorry. I'm a bit confused. I just realised the question does not even have a square root, lol. Sorry!

    • 2 years ago
  16. lalaly Group Title
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    haha so what is the question?

    • 2 years ago
  17. need_help! Group Title
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    I have just edited the question.

    • 2 years ago
  18. eliassaab Group Title
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    \[ (4x^2-3)^2=64 \\ (4x^2-3)=8 \\ 4 x^2 = 11\\ x^2 = \frac {11}4\\ x=\pm \sqrt{\frac {11}4}\\ x =\pm \frac {\sqrt {11}} 2 \]

    • 2 years ago
  19. need_help! Group Title
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    Thank you so much =)

    • 2 years ago
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