# Thread: If (x + 2/x)^2 = 36, then what is (x^2 + 2/x^2)^2?

1. ## If (x + 2/x)^2 = 36, then what is (x^2 + 2/x^2)^2?

Can any kind soul help me in the below?

a) Factorise t^4 - 16
b) If (x + 2/x)^2 = 36, then what is (x^2 + 2/x^2)^2?

2. Well, a) is easy, if you note that $16 = 4^2$ (or $2^4$).

Second question : I guess there is another more clever way to do this but I would have solved for $x$ first, then evaluated.

Since $(x + \frac{2}{x})^2 = 36$, we have $x + \frac{2}{x} = 6$, thus $\frac{x^2}{x} + \frac{2}{x} = 6$, therefore $\frac{x^2 + 2}{x} = 6$, so $x^2 + 2 = 6x$, and so, $x^2 - 6x + 2 = 0$.

By means of the quadratic formula (for example), you find that $x$ equals that value and that value, then you substitute your results back into the second equation and evaluate.

Does that make sense ?

3. Dear Bacterius, thanks so much!! but i cannot get the answer of 32.
The topic is solving quadratic equations by factorisation.

4. Originally Posted by ppppp77
Can any kind soul help me in the below?

a) Factorise t^4 - 16

Remember that $a^2-b^2=(a-b)(a+b)$ , so in this case substitute $a=t^2\,,\,\,b=4$ ...

b) If (x + 2/x)^2 = 36, then what is (x^2 + 2/x^2)^2?
We have that $\left(\frac{x+2}{x}\right)^2=36\Longrightarrow \frac{x+2}{x}=\pm 6\Longrightarrow$ ....get the two possible values for x and substitute.

Tonio

5. Originally Posted by ppppp77
b) If (x + 2/x)^2 = 36, then what is (x^2 + 2/x^2)^2?
wonder if in the second expression it's actually a $\frac4{x^2}$ instead of the $\frac2{x^2},$ because $\left( x^{2}+\frac{4}{x^{2}} \right)^{2}=\left( \left( x+\frac{2}{x} \right)^{2}-4 \right)^{2}=32^{2}.$

6. Thanks so much Krizalid! I thought the same too.