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Aadarsh
The number of real roots of (x+3)^4 + (x+5)^4 = 16 is: a) 0 b) 2 c) 4 d) none of these
Please say the detailed steps.
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
@experimentX can u explain the direct solution and step? I didn't get it.
|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)
But should we do the graph for solving this? Is der no direct method?
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
@Aadarsh you can solve also in this way..
Its ok. But without graph, I want to solve this. please say some other way.
-5 and -3 are quite visible .. so factorize it (x+3)(x+5)(some terms ..) = 0 ... first find some terms
can you simplify this equation? (x+3)^4 + (x+5)^4 - 16 =0
Not getting exactly. But can we apply Neumark and Ferrari-Langrange method for solving?
well i guess you can
\[(x+3)^4+(x+5)^4-16=0 \]\[2 (x+3) (x+5) \left(x^2+8 x+23\right)=0 \]
well, that what i mean ..!
Oh yeah, I got the factorisation.
(x2+8x+23) will give you two complex roots, so your real roots are only -5 and -3
Thank you for the medals.
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..
I have not read those concepts. Just completed grade 10. So facing problems.
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..
Ji haan, I solved using trial method, got -5 and -3.
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..
Really? Yeah preparing for IIT