0

I am stuck on the following problem. I also mean to say greet you with a Hello! but for some reason it won't save it haha.

This problem was part of my homework, which is now past due and I have received partial credit for. The answer was released and I am trying to understand it.

I am completely stuck on the "Solving the equation $1+ \frac{2\sqrt{21}}{9} = (3a+a\sqrt{21})^3 $ for $a$" part. No matter how I break it down, I can't get that answer. Also, it seems like the $a=-6$ as the coefficient of the $x^2$ term has nothing to do with this part of the problem. Can someone explain please?

enter image description here

mathjohnn
  • 181
  • 11

1 Answers1

2

It is confusing for them to use $a$ in the hint, since it coincides with the coefficient of the $x^2$ term, and has nothing to do with the purpose of the hint.

Now the hint was

\begin{align}1+ \frac{2\sqrt{21}}{9} &= (3a+a\sqrt{21})^3 \\&=a^3(3+\sqrt{21})^3 \\&= a^3(3^3+3\cdot3^2\sqrt{21}+3\cdot3\sqrt{21}^2+\sqrt{21}^3) \\&=a^3(27+27\sqrt{21}+189+21\sqrt{21}) \\&=a^3(216+48\sqrt{21}) \\\leadsto a^3=\frac 19\frac{9+2\sqrt{21}}{216+48\sqrt{21}}&=\frac19\frac1{24}=\frac1{216}\end{align} (Factor the denominator and the numerator appears so a cancellation takes place)

which gives $a=\dfrac16$. Hence $1+ \dfrac{2\sqrt{21}}{9} = (3a+a\sqrt{21})^3 = \left(\dfrac12+\dfrac16\sqrt{21}\right)^3$.

This hint is only used to simplify the root

$$u=\sqrt[3]{1+ \frac{2\sqrt{21}}{9}}$$

imranfat
  • 10,029
player3236
  • 16,413
  • sharp approach! – imranfat Nov 06 '20 at 03:42
  • @imranfat What do you mean by "A conjugate is used to take out the radical in the numerator"? – player3236 Nov 06 '20 at 03:44
  • thank you so much! I honestly would never have approached it that way. Did in a few lines what I was trying to understand for a couple hours! – mathjohnn Nov 06 '20 at 03:59
  • 1
    @player3236 "What do you mean by : A conjugate is used to take out the radical in the numerator?" He means that $$\frac{9 + 2\sqrt{21}}{216 + 48\sqrt{21}} \times \frac{9 - 2\sqrt{21}}{9 - 2\sqrt{21}} ~=~ \frac{-3}{1944 + 432\sqrt{21} - 432\sqrt{21} - 2016}.$$ – user2661923 Nov 06 '20 at 05:55
  • @user2661923 I certainly did not approach it that way. I simply observed that $216+48\sqrt{21} = 24(9+2\sqrt {21})$. – player3236 Nov 06 '20 at 05:57
  • Interesting situation. First of all, I didn't notice the point that you are making, which is certainly the best approach. Second of all, I got confused - I assumed (wrongly) that you were not the person who created the answer and that you were questioning what a specific line in the answer intended. So, I assumed that you were the OP, and simply got out my calculator to answer. My bad. – user2661923 Nov 06 '20 at 06:01
  • @user2661923 A perplexing situation indeed (usually people point out my mistakes/suggest improvements in the comments). I know the radicals can be eliminated since it was given as a hint. – player3236 Nov 06 '20 at 06:03
  • Of course, I see it now. Simple factoring (I edited). I did it with a conjugate, but obviously simply factoring is much better – imranfat Nov 06 '20 at 13:49