anonymous
  • anonymous
The number of real roots of (x+3)^4 + (x+5)^4 = 16 is: a) 0 b) 2 c) 4 d) none of these
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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experimentX
  • experimentX
2
experimentX
  • experimentX
-3 and -5
anonymous
  • anonymous
Please say the detailed steps.

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experimentX
  • experimentX
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
anonymous
  • anonymous
@experimentX can u explain the direct solution and step? I didn't get it.
anonymous
  • anonymous
|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)
anonymous
  • anonymous
But should we do the graph for solving this? Is der no direct method?
experimentX
  • experimentX
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
anonymous
  • anonymous
@Aadarsh you can solve also in this way..
anonymous
  • anonymous
Its ok. But without graph, I want to solve this. please say some other way.
experimentX
  • experimentX
-5 and -3 are quite visible .. so factorize it (x+3)(x+5)(some terms ..) = 0 ... first find some terms
anonymous
  • anonymous
Can u do and say?
experimentX
  • experimentX
can you simplify this equation? (x+3)^4 + (x+5)^4 - 16 =0
anonymous
  • anonymous
Not getting exactly. But can we apply Neumark and Ferrari-Langrange method for solving?
experimentX
  • experimentX
well i guess you can
anonymous
  • anonymous
\[(x+3)^4+(x+5)^4-16=0 \]\[2 (x+3) (x+5) \left(x^2+8 x+23\right)=0 \]
experimentX
  • experimentX
well, that what i mean ..!
anonymous
  • anonymous
Oh yeah, I got the factorisation.
experimentX
  • experimentX
(x2+8x+23) will give you two complex roots, so your real roots are only -5 and -3
anonymous
  • anonymous
Thank you for the medals.
anonymous
  • anonymous
Most welcome
anonymous
  • anonymous
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..
anonymous
  • anonymous
I have not read those concepts. Just completed grade 10. So facing problems.
anonymous
  • anonymous
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..
anonymous
  • anonymous
Ji haan, I solved using trial method, got -5 and -3.
anonymous
  • anonymous
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..
anonymous
  • anonymous
Really? Yeah preparing for IIT

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