If x2-8x-1=0 then prove that x2+1/x2 ?

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Answer 1 · 2018-06-21T14:20:01

#x^2 + 1/x^2 = (64x^2 + 16x +2)/ (1+8x)#

#x^2-8x-1=0#

To Find: #x^2+1/x^2#
Solution:
#x^2-8x-1=0#

#=> x^2 = 1+8x #

#=> therefore 1/x^2 =1/ (1 + 8x)#

#=> therefore x^2 + 1/x^2 = (1+8x) + 1/ (1+8x) #

#=> x^2 + 1/x^2 = ((1+8x)(1+8x) + 1)/ (1+8x) #

#= ((1+8x)^2 + 1)/ (1+8x) = (1 + 16x + 64x^2 +1)/ (1+8x)#

#= (64x^2 + 16x +2)/ (1+8x)#

So, #x^2 + 1/x^2 = (64x^2 + 16x +2)/ (1+8x)#

Answer 2 · 2018-06-21T15:07:48

The question looks incomplete, but I'll do whatever I can

Since we know that

#x^2-8x-1=0#

we can isolate #x^2# to get

#x^2=8x+1#

So, #x^2+1/x^2# becomes

#(8x+1)+1/(8x+1) = ((8x+1)^2+1)/(8x+1) = \frac{64x^2+16x+2}{8x+1}#

Now, I don't know what you want to prove, but for sure we have

#x^2+1/x^2=\frac{64x^2+16x+2}{8x+1}#

Answer 3 · 2018-06-22T10:42:23

# x^2+1/x^2=66#.

#"Given that, "x^2-8x-1=0............(ast)#.

#"Here, "x!=0, because, "if "x=0," then, "#

#x^2-8x-1=0 rArr 0-1=0," which is not possible."#

#"Hence, dividing "(ast)" throughout by "x!=0," we get, "#

#x-8-1/x=0, or, x-1/x=8#.

#:. (x-1/x)^2=8^2=64#.

#:. x^2-2+1/x^2=64#.

# rArr x^2+1/x^2=64+2=66#.