Question

If \(\LARGE{x-\frac{1}{x}=\sqrt{21}}\), then

\(\Huge{\left( x^2+\frac{1}{x^2} \right) \left( x+\frac{1}{x} \right)=?}\)

Answer 1

I did this through right here

\[\LARGE{\left( x-{1 \over x} \right)^2=(\sqrt{21})^2}\]\[\LARGE{x^2+\frac{1}{x^2}-2=21}\]\[\LARGE{x^2+\frac{1}{x^2}=23}\]then i put it's value in the equation\[\LARGE{(23)\left( x+\frac{1}{x} \right)}\]My real doubt is how to find the value of \(\Large{x+\frac{1}{x}}\).

Answer 2

hint : (x+1/x)^2 = x^2+1/x^2 +2

Answer 3

You can use the fact that:\[\left(x+\frac{1}{x}\right)^2=x^2+\frac{1}{x^2}+2\]

Answer 4

and you know x^2+1/x^2=...

Answer 5

Ohh ! ok thanx i gt it :)