# Given that (a-1/a)=3 determine the value of (a+1/a)^2 without solving for x

Please excuse my noobery but I'm very new to high school maths for reasons unimportant.

I'm using this site as my study guide to get through high school math and the questions (close to the bottom of the page) that I'm having trouble with go like this:

(a) Expand: $(a+\frac{1}{a})^2$.

(b) Given that $(a+\frac{1}{a})=3$, determine the value of $(a+\frac{1}{a})^2$ without solving for $x$.

(c) Given that $(a-\frac{1}{a})=3$, determine the value of $(a+\frac{1}{a})^2$ without solving for $x$.

I can do (a) and (b) but I get stuck at (c) even though the website gives the answers but their explanation doesn't make any sense and I think the problem is that the site doesn't explain how to do these sorts of problems, what these problems are about, where to go find out what all this is, etc. so I'm hoping someone can please explain to me how to do this step for step so that my puny mind can understand it too.

Edit: If you'd like me to post the answers according to the website, please let me know, I only exclude them because of how much space that would take.

• $$\left(a+\dfrac1a\right)^2=?$$ and $$\left(a-\dfrac1a\right)^2=?$$ – lab bhattacharjee May 21 '17 at 15:23
• for $(b)$ it is simply 9. – hamam_Abdallah May 21 '17 at 15:24
• Oh right, I actually got that one too, lemme quickly update. – PrintlnParams May 21 '17 at 15:25
• Where is the variable $x$ used? – Jose Arnaldo Bebita-Dris May 21 '17 at 15:26

hint just observe that

$$(a+1/a)^2=(a-1/a)^2+4$$ because

$$(a+1/a)^2=$$ $$a^2+1/a^2+2.a.(1/a)=$$ $$a^2+1/a^2+2$$ and $$(a-1/a)^2=$$ $$a^2+1/a^2-2.a. (1/a) =$$ $$a^2+1/a^2-2=$$ $$a^2+1/a^2+2 -4$$

• @JoseArnaldoBebitaDris It is obvious that we can make mistake. Thanks. – hamam_Abdallah May 21 '17 at 15:29
• Thanks but how do we know that adding 4 will change the plus sign to a minus? So for example why couldn't I add 8 or subtract 2 or multiply by 432, etc. to get change the positive to a negative? – PrintlnParams May 21 '17 at 15:29

(b) If $\displaystyle a+\frac1a=3$, then $\displaystyle\left(a+\frac1a\right)^2=3^2=9$.

(c) If $\displaystyle a-\frac1a=3$ then $\displaystyle\left(a-\frac1a\right)^2=9$. But $\displaystyle\left(a-\frac1a\right)^2=a^2-2+\frac1{a^2}$ and so $\displaystyle\left(a+\frac1a\right)^2=a^2+2+\frac1{a^2}=\left(a^2-2+\frac1{a^2}\right)+4=13.$

For (a) you have to use this:

$(x+y)^2= x^2+y^2 +2xy$

For (b) if you have:

$x=3$ then $x^2 = 3^2$. If you substitute $x$ with the value $(a+1/a)$ then $9=3^2=x^2=(a+1/a)^2$

For (c) you can use the hint given by Salahamam_Fatima and the (b) section, now for $x=(a-1/a)$. Can you solve this now?