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

3 Answers
Jun 21, 2018

x^2 + 1/x^2 = (64x^2 + 16x +2)/ (1+8x)x2+1x2=64x2+16x+21+8x

Explanation:

x^2-8x-1=0x28x1=0

To Find: x^2+1/x^2x2+1x2
Solution:
x^2-8x-1=0x28x1=0

=> x^2 = 1+8x x2=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)

Jun 21, 2018

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

Explanation:

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}

Jun 22, 2018

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

Explanation:

"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.