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

- 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

- chestercat

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- experimentX

2

- experimentX

-3 and -5

- anonymous

Please say the detailed steps.

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## More answers

- 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

@experimentX can u explain the direct solution and step? I didn't get it.

- 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

But should we do the graph for solving this? Is der no direct method?

- 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

@Aadarsh you can solve also in this way..

- anonymous

Its ok. But without graph, I want to solve this. please say some other way.

- experimentX

-5 and -3 are quite visible .. so factorize it
(x+3)(x+5)(some terms ..) = 0 ... first find some terms

- anonymous

Can u do and say?

- experimentX

can you simplify this equation?
(x+3)^4 + (x+5)^4 - 16 =0

- anonymous

Not getting exactly. But can we apply Neumark and Ferrari-Langrange method for solving?

- experimentX

well i guess you can

- anonymous

\[(x+3)^4+(x+5)^4-16=0 \]\[2 (x+3) (x+5) \left(x^2+8 x+23\right)=0 \]

- experimentX

well, that what i mean ..!

- anonymous

Oh yeah, I got the factorisation.

- experimentX

(x2+8x+23) will give you two complex roots, so your real roots are only -5 and -3

- anonymous

Thank you for the medals.

- anonymous

Most welcome

- 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

I have not read those concepts. Just completed grade 10. So facing problems.

- 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

Ji haan, I solved using trial method, got -5 and -3.

- 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

Really? Yeah preparing for IIT

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