If $a+\frac1a=\sqrt3$ then $a^4+\frac1{a^4}=\ ?$

Question

If $a+\frac1a=\sqrt3$ then $a^4+\frac1{a^4}=\ ?$

Can someone please explain to me how to solve this? because I tried everything I know and it didn't work.

P.S: I'm in 8th grade so no quadratic formula.

Answer 1

We have these identities. (Do you see why these are true? Hint: FOIL. Or, if you learned $``{(a+b)^2=a^2+2ab+b^2} ``$, use that.)

$$\left(a+\frac1a\right)^2=\left(a^2+\frac1{a^2}\right)+2$$

$$\left(a^2+\frac1{a^2}\right)^2=\left(a^4+\frac1{a^4}\right)+2$$

Now, plug in $(a+1/a)=\sqrt3$.

$$\left(\sqrt3\right)^2=\left(a^2+\frac1{a^2}\right)+2$$

So $3=(a^2+1/a^2)+2$, and $a^2+1/a^2=1$.

$$(1)^2=\left(a^4+\frac1{a^4}\right)+2$$

So $1=(a^4+1/a^4)+2$, or $a^4+1/a^4=-1$ and we have solved the problem.

Answer 2

Apply $p^2+q^2=(p+q)^2-2pq$

$$a^4+\frac1{a^4}=\left(a^2\right)^2+\left(\frac1{a^2}\right)^2$$

and on $$a^2+\frac1{a^2}=a^2+\left(\dfrac1a\right)^2$$

Answer 3

$$3=\left(a+\frac 1a\right)^2=a^2+\frac1{a^2}+2$$ so $$1=a^2+\frac1{a^2}$$

Make the same manipulation again and you get it.

But it seems that all that is impossible for a real number $a$...

Answer 4

$$\left(a^n+\frac1{a^n}\right)\left(a+\frac1a\right)=\left(a^{n+1}+\frac1{a^{n+1}}\right)+\left(a^{n-1}+\frac1{a^{n-1}}\right)$$

$$\implies a^{n+1}+\frac1{a^{n+1}}=\left(a^n+\frac1{a^n}\right)\left(a+\frac1a\right)-\left(a^{n-1}+\frac1{a^{n-1}}\right)$$

If $T_m=a^m+\dfrac1{a^m},$

$$T_{n+1}=T_n\cdot T_1-T_{n-1}$$

We have $T_1=\sqrt3,T_0=2$ as $a\ne0$

Answer 5

Expand $\left(a + \frac{1}{a}\right)^4$ by using foil, repeatedly using the famous result $(a+b)^2 = a^2 + 2ab + b^2$ (which is best in this case) or as I've done, you can use the binomial theorem assisted by pascal's triangle: $$\left(a + \frac1{a}\right)^4 = \left(\sqrt{3}\right)^4\\ \\ a^4 + 4\cdot a^3\frac{1}{a} + 6\cdot a^2\left(\frac{1}{a}\right)^2 + 4\cdot a\left(\frac{1}{a}\right)^3 + \left(\frac{1}{a}\right)^4 = 3^{4/2}\\ a^4 + 4a^2 + 6 + \frac{4}{a^2} + \frac{1}{a^4} = 3^2\\ a^4 + \frac{1}{a^4} + 4\left(a^2 + \frac{1}{a^2}\right) = 9 - 6\\ a^4 + \frac{1}{a^4} + 4\left(\Big(a + \frac{1}{a}\Big)^2 - 2\, a\frac{1}{a}\right) = 3\\ a^4 + \frac{1}{a^4} + 4\left((\sqrt{3})^2 - 2\right) = 3\\ \therefore \quad a^4 + \frac{1}{a^4} = 3 - 4\cdot(3-2) = -1 $$ What you see above is a gut response to this question, if you've paid attention, you will notice my usage of $a^2 + b^2 = (a+b)^2 - 2ab$ and the important substitution of $a + \frac{1}{a} = \sqrt{3}$

There are loads of ways to do algebra. The results matter more than the method. Intuition matters more than anything else.