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Aadarsh Group Title

The number of real roots of (x+3)^4 + (x+5)^4 = 16 is: a) 0 b) 2 c) 4 d) none of these

  • 2 years ago
  • 2 years ago

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  1. experimentX Group Title
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    2

    • 2 years ago
  2. experimentX Group Title
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    -3 and -5

    • 2 years ago
  3. Aadarsh Group Title
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    Please say the detailed steps.

    • 2 years ago
  4. experimentX Group Title
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    here's an ugly method http://math.stackexchange.com/questions/785/is-there-a-general-formula-for-solving-4th-degree-equations simplify it to 4th degree equation ... it has at max 4 roots, in our case it's 2 real and 2 complex

    • 2 years ago
  5. Aadarsh Group Title
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    @experimentX can u explain the direct solution and step? I didn't get it.

    • 2 years ago
  6. Taufique Group Title
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    |dw:1332919098675:dw| the no of solution =cut points of graphs(y=(x+3)^4 and y=16-(x+5)^4)=2 (they are -3 and -5)

    • 2 years ago
  7. Aadarsh Group Title
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    But should we do the graph for solving this? Is der no direct method?

    • 2 years ago
  8. experimentX Group Title
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    well i must say, @Taufique is quite correct in explaining above ... since curve cuts at two places, it has two real solutions, and two complex solutions

    • 2 years ago
  9. Taufique Group Title
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    @Aadarsh you can solve also in this way..

    • 2 years ago
  10. Aadarsh Group Title
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    Its ok. But without graph, I want to solve this. please say some other way.

    • 2 years ago
  11. experimentX Group Title
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    -5 and -3 are quite visible .. so factorize it (x+3)(x+5)(some terms ..) = 0 ... first find some terms

    • 2 years ago
  12. Aadarsh Group Title
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    Can u do and say?

    • 2 years ago
  13. experimentX Group Title
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    can you simplify this equation? (x+3)^4 + (x+5)^4 - 16 =0

    • 2 years ago
  14. Aadarsh Group Title
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    Not getting exactly. But can we apply Neumark and Ferrari-Langrange method for solving?

    • 2 years ago
  15. experimentX Group Title
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    well i guess you can

    • 2 years ago
  16. robtobey Group Title
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    \[(x+3)^4+(x+5)^4-16=0 \]\[2 (x+3) (x+5) \left(x^2+8 x+23\right)=0 \]

    • 2 years ago
  17. experimentX Group Title
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    well, that what i mean ..!

    • 2 years ago
  18. Aadarsh Group Title
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    Oh yeah, I got the factorisation.

    • 2 years ago
  19. experimentX Group Title
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    (x2+8x+23) will give you two complex roots, so your real roots are only -5 and -3

    • 2 years ago
  20. robtobey Group Title
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    Thank you for the medals.

    • 2 years ago
  21. Aadarsh Group Title
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    Most welcome

    • 2 years ago
  22. Taufique Group Title
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    First of all , find the dY/dX of the given curve.dy/dx means tangent on the curve. find the cut point of dy/dx and X axis on putting Y=0 in the equation ,and solve it it gives you solution of the equation..

    • 2 years ago
  23. Aadarsh Group Title
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    I have not read those concepts. Just completed grade 10. So facing problems.

    • 2 years ago
  24. Taufique Group Title
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    ok Aadarsh ji ,i thought that you are in 12 th, you can solve it using hit and trial method.. and factorise this after it..

    • 2 years ago
  25. Aadarsh Group Title
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    Ji haan, I solved using trial method, got -5 and -3.

    • 2 years ago
  26. Taufique Group Title
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    If you will prepare for iit then you get a complex equation and on that situation it is very tough to guess what is the root...then there is many method to solve such a problem..

    • 2 years ago
  27. Aadarsh Group Title
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    Really? Yeah preparing for IIT

    • 2 years ago
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