Besides 0, what are the answers for 'a' in this equation?? Watch

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Trivial Questio
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#1
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What does 'a' equal?

a + square root (a^2 + square root(a^3 + 1)) = 1

*Note: I couldn't find a square root sign, so just bear with it.
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arxtra
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#2
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well I end up with

a(a-2)²=0

so I guess you end up with a = 0, 2 ?

(Original post by Trivial Questio)
What does 'a' equal?

a + square root (a^2 + square root(a^3 + 1)) = 1

*Note: I couldn't find a square root sign, so just bear with it.
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J.F.N
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(Original post by Trivial Questio)
What does 'a' equal?

a + square root (a^2 + square root(a^3 + 1)) = 1

*Note: I couldn't find a square root sign, so just bear with it.
Is there a restriction on a? If you're only considering integer values, then 0 is the only solution. (Because rt(a^2 +rt(a^3 +1)) is a non-integer for all integers a, and you can't have an integer + a non-integer = an integer. This, remarkably, relates to Catalan's theorem, because in order for the rt(a^2 +rt(a^3 +1)) term to be an integer, then we need a^3 +1 to be a perfect square, and the only (nontrivial) case this happens, as Catalan's theorem asserts, is for a=2... which does NOT work anyway).

Edit: In fact, 0 is the only solution. Consider a>0. Clearly, a cannot be greater than 1. So we have 0<a=<1 . Now, since a>0, then rt(a^2 +rt(a^3 +1) >0. You can easily see that as a-->0, rt(a^2 +rt(a^3 +1) -->1. So a=0 is the only solution.

Similarly, consider a<0. a cannot be smaller than -1, because if it were, then rt(a^3 + 1) is undefined. So -1<a<0. Now in a similar way, as a-->0, rt(a^2 +rt(a^3 +1)-->1. So a=0 is the only solution.
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Trivial Questio
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'a' has to be a real integer, so I suppose the only real answer is 0. thanks.
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J.F.N
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(Original post by Trivial Questio)
'a' has to be a real integer, so I suppose the only real answer is 0. thanks.
Yep, see the edited above. Even if a is a real number (not necessarily an integer), a=0 is the only solution.
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RichE
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(Original post by Newton)
sqrt[(a^2)+sqrt[(a^3)+1]]=1-a

=>(a^2)+sqrt[(a^3)+1]=1-2a+(a^2)

=>sqrt[(a^3)+1]=1-2a

=>(a^3)+1=1-4a+4(a^2)

=>(a^3)-4(a^2)+4a=0

=>a((a^2)-4a+4)=0

=>a(a-2)(a-2)=0

=>a=0 OR a=2

Newton.
Have you tried putting 2 back in the original equation?

squaring up equations can add extra roots
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J.F.N
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(Original post by Newton)
sqrt[(a^2)+sqrt[(a^3)+1]]=1-a

=>(a^2)+sqrt[(a^3)+1]=1-2a+(a^2)

=>sqrt[(a^3)+1]=1-2a

=>(a^3)+1=1-4a+4(a^2)

=>(a^3)-4(a^2)+4a=0

=>a((a^2)-4a+4)=0

=>a(a-2)(a-2)=0

=>a=0 OR a=2

Newton.
I didn't look over the working, but I'm sure there's something wrong. a=0 is the only solution. Plug in 2, you get 2+rt(7)=1.
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RichE
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(Original post by RichE)
Have you tried putting 2 back in the original equation?

squaring up equations can add extra roots
as I said squaring can add extra roots

x=1 implies x^2 = 1

but their roots are different as the second includes x = -1

all one can say is that the roots of the first are a subset of the roots of the latter
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yazan_l
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Originally Posted by Newton
sqrt[(a^2)+sqrt[(a^3)+1]]=1-a

=>(a^2)+sqrt[(a^3)+1]=1-2a+(a^2)

=>sqrt[(a^3)+1]=1-2a

=>(a^3)+1=1-4a+4(a^2)

=>(a^3)-4(a^2)+4a=0

=>a((a^2)-4a+4)=0

=>a(a-2)(a-2)=0

=>a=0 OR a=2

Newton.

2 + sqrt (2² + sqrt(2³ + 1)) =
2 + sqrt (4 + sqrt(9)) =
2 ± sqrt ( 4 ± 3 )
you will have four values for this:
4.64 , -0.65, 1 , 3
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Christophicus
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#10
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Here's the graph for the function.
Attached files
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RichE
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#11
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(Original post by yazan_l)
2 + sqrt (2² + sqrt(2³ + 1)) =
2 + sqrt (4 + sqrt(9)) =
2 ± sqrt ( 4 ± 3 )
you will have four values for this:
4.64 , -0.65, 1 , 3
If one just writes sqrt(x) or \sqrt x then this denotes the positive square root

That's the reason for the plus or minus in the quadratic formula
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