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burhan101

Solve for 'x'

  • 10 months ago
  • 10 months ago

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  1. burhan101
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    \[\huge 0=1- \frac{ 1 }{ 2x^{\frac{ 3 }{ 2}}} \]

    • 10 months ago
  2. Abhishek619
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    |dw:1371705222820:dw|

    • 10 months ago
  3. burhan101
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    how does it equal one ?!

    • 10 months ago
  4. terenzreignz
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    slowly :D Why don't we bring \(\huge \frac1{2x^{\frac32}}\) to the left side like this... \[\Large \frac1{2x^{\frac32}}=1\] clear now? :)

    • 10 months ago
  5. burhan101
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    |dw:1371743475441:dw|

    • 10 months ago
  6. burhan101
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    yes

    • 10 months ago
  7. terenzreignz
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    We basically added \[\huge \frac1{2x^{\frac32}}\] to both sides as per the addition property of equality.

    • 10 months ago
  8. terenzreignz
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    Well then, can you solve it from here?

    • 10 months ago
  9. burhan101
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    do i multiply by both sides?

    • 10 months ago
  10. terenzreignz
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    Sure :)

    • 10 months ago
  11. burhan101
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    |dw:1371743682268:dw|

    • 10 months ago
  12. terenzreignz
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    Never mind that for now. Why don't we multiply both sides by \[\Large 2x^{\frac32}\] That gives us... \[\Large 1 = 2x^{\frac32}\] And divide both sides by 2...

    • 10 months ago
  13. burhan101
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    \[x^{3/2} =1/2 \]

    • 10 months ago
  14. terenzreignz
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    true dat :) Now you can raise both sides to the \(\LARGE \frac23\) power (just to get rid of the exponent of x) You then get...?

    • 10 months ago
  15. burhan101
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    |dw:1371743929750:dw| like this?

    • 10 months ago
  16. terenzreignz
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    That gives you x. What about the right-hand side?

    • 10 months ago
  17. burhan101
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    0/63

    • 10 months ago
  18. burhan101
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    0.63 *

    • 10 months ago
  19. terenzreignz
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    come again? :D

    • 10 months ago
  20. terenzreignz
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    Oh, were you supposed to give it in decimal form? okay then :D

    • 10 months ago
  21. burhan101
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    I can get a fraction answer too?

    • 10 months ago
  22. terenzreignz
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    Well, the best you can hope for is the ugly-looking \[\Large \left(\frac12\right)^{\frac23}\]

    • 10 months ago
  23. terenzreignz
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    and that is the correct answer :)

    • 10 months ago
  24. burhan101
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    oh so i should probably keep it,

    • 10 months ago
  25. burhan101
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    thank youuu !!!

    • 10 months ago
  26. terenzreignz
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    Yeah, unless you were specifically asked for an approximation :)

    • 10 months ago
  27. terenzreignz
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    Oh well... now I have to go :) Please practice solving algebraic equations :D ----------------------------------------- Terence out

    • 10 months ago
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