Roots from perfect squares

[mathwannabe]

Here is the problem:

1) $\displaystyle \sqrt{|12\sqrt{5}-29|}-\sqrt{12\sqrt{5}+29}=$ ?

As this is a problem for some of the previous exams for getting into faculty (and everything has to be neat x)) I figured that $\displaystyle |12\sqrt{5}-29|$ and $\displaystyle 12\sqrt{5}+29$ must be perfect squares of something.

So, I got to :

$\displaystyle \sqrt{|-(2\sqrt{5}-3)^2|}-\sqrt{(2\sqrt{5}+3)^2}=$

$\displaystyle =\sqrt{(2\sqrt{5}-3)^2}-\sqrt{(2\sqrt{5}+3)^2}=$

$\displaystyle =(2\sqrt{5}-3)-(2\sqrt{5}+3)=$

$\displaystyle =2\sqrt{5}-3-2\sqrt{5}-3)=$

$\displaystyle =-6$

I realized that the point of this problem is not to recognize those perfect squares, but to deal with absolute in the first term.
I figured that the only way to get a $\displaystyle 12\sqrt{5}-29$ from squaring $\displaystyle (2\sqrt{5}-3)$ was to put that minus sign in front of that bracket, which becomes a plus when I remove the absolute brackets (I don't know if terminology is correct, English is not my main language, I am from Serbia). As this problem caused a lot of confusion for me, I am unsure if my result is correct, so, please help

By the way, is there some bullet proof method for finding a root of a perfect polynomial square if it contains roots? Or it just comes to thinking "hard"?

 

[pka]

I would rewrite $\displaystyle \sqrt {\left| {12\sqrt 5 - 29} \right|} = \sqrt {29 - 12\sqrt 5 }$ without absolute value.
Then square the expression to get
$\displaystyle \left( {29 - 12\sqrt 5 } \right) - 22 + \left( {29 + 12\sqrt 5 } \right)$.
Now remember the original is negative.

 

[mathwannabe]

I would rewrite $\displaystyle \sqrt {\left| {12\sqrt 5 - 29} \right|} = \sqrt {29 - 12\sqrt 5 }$ without absolute value.
Then square the expression to get
$\displaystyle \left( {29 - 12\sqrt 5 } \right) - 22 + \left( {29 + 12\sqrt 5 } \right)$.
Now remember the original is negative.

I really don't understand this, must be one of the holes in my knowledge I'm dragging with me.

How did you get loose of those absolute brackets? By simply changing the sign of each term within the brackets?

Also, how did you square the expression to get $\displaystyle \left( {29 - 12\sqrt 5 } \right) - 22 + \left( {29 + 12\sqrt 5 } \right)$?

Could you please provide me with a more in-depth analysis of this, It is really important for me to understand this matter and your help would be much appreciated.

 

[pka]

How did you get loose of those absolute brackets? By simply changing the sign of each term within the brackets?

Because $\displaystyle 12\sqrt5-29<0$ we have $\displaystyle |12\sqrt5-29|=29-12\sqrt5$

Also, how did you square the expression to get $\displaystyle \left( {29 - 12\sqrt 5 } \right) - 22 + \left( {29 + 12\sqrt 5 } \right)$?

$\displaystyle \left(\sqrt{a-b}-\sqrt{a+b}\right)^2=(a-b)-2\sqrt{a^2-b^2}+(a+b)$

 

[mathwannabe]

Because $\displaystyle 12\sqrt5-29<0$ we have $\displaystyle |12\sqrt5-29|=29-12\sqrt5$


$\displaystyle \left(\sqrt{a-b}-\sqrt{a+b}\right)^2=(a-b)-2\sqrt{a^2-b^2}+(a+b)$

Yes, I understand now how to get rid of the absolute brackets and how to square the $\displaystyle 29-12\sqrt5$ but I don't understand why would I square it because I have to find the root expression so I could cancel the root and the square. Here is what I mean:


$\displaystyle \sqrt{|12\sqrt{5}-29|}-\sqrt{12\sqrt{5}+29}=$

$\displaystyle =\sqrt{29-12\sqrt5}-\sqrt{12\sqrt{5}+29}=$

$\displaystyle =\sqrt{(3-2\sqrt5)^2}-\sqrt{(2\sqrt{5}+3)^2}=$

$\displaystyle =(3-2\sqrt5)-(2\sqrt{5}+3)=$

$\displaystyle =3-2\sqrt5-2\sqrt{5}-3=$

$\displaystyle =-4\sqrt5$

Is this correct now?

 

[pka]

$\displaystyle \sqrt{|12\sqrt{5}-29|}-\sqrt{12\sqrt{5}+29}=$

$\displaystyle =\sqrt{29-12\sqrt5}-\sqrt{12\sqrt{5}+29}=$

$\displaystyle =\sqrt{(3-2\sqrt5)^2}-\sqrt{(2\sqrt{5}+3)^2}=$

NO, NO, & NO-NO.
Square whole expression:
$\displaystyle \left(\sqrt{29-12\sqrt5}-\sqrt{12\sqrt{5}+29}\right)^2$
$\displaystyle =29-12\sqrt5+22+29+12\sqrt5=36$
So the negative answer is $\displaystyle -\sqrt{36}=-6$

 

[soroban]

Hello, mathwannabe!

Why did they write it that way? . . . Silly!

\(\displaystyle \sqrt{29 -12\sqrt{5}}-\sqrt{29 + 12\sqrt{5}\)

Your game plan is a good one.
I'd solve it like this . . .

Let $\displaystyle a + b\sqrt{5} \:=\:\sqrt{29 + 12\sqrt{5}}$ . where $\displaystyle a$ and $\displaystyle b$ are rational.

Then: .$\displaystyle (a + b\sqrt{5})^2 \:=\:29 + 12\sqrt{5} \quad\Rightarrow\quad a^2 + 2ab\sqrt{5} + 5b^2 \:=\:29 + 12\sqrt{5}$

We have: .$\displaystyle (a^2 + 5b^2) + (2ab)\sqrt{5} \;=\;29 + 12\sqrt{5}$

Equate rational and irrational components: .$\displaystyle \begin{Bmatrix} a^2 + 5b^2 \:=\:29 & [1] \\ 2ab \:=\:12 & [2] \end{Bmatrix}$

From [2]: .$\displaystyle b \,=\,\frac{6}{a}$

Substitute into [1]: .$\displaystyle a^2 + 5\left(\frac{6}{a}\right)^2 \:=\:29 \quad\Rightarrow\quad a^2 + \frac{180}{a^2} \:=\:29$

. . $\displaystyle a^4 + 180 \:=\:29a^2 \quad\Rightarrow\quad a^4 - 29a^2 + 180 \:=\:0 \quad\Rightarrow$ . . $\displaystyle (a^2 - 9)(a^2-20)\:=\:0$

We have: .\(\displaystyle \begin{array}{ccccccccccccc}a^2 - 9 \:=\: 0 & \Rightarrow & a^2 \:=\:9 & \Rightarrow & a \:=\: 3,\;b\:=\: 2 \\
a^2-20 \:=\:0 & \Rightarrow & a^2 \:=\:20 & \Rightarrow & \text{Irrational} \end{array}\)


Therefore: .$\displaystyle \sqrt{29 - 12\sqrt{5}} - \sqrt{29+12\sqrt{5}} \:=\;\sqrt{(3-2\sqrt{5})^2} - \sqrt{(3 + 2\sqrt{5})^2}$

. . . . . . . . .$\displaystyle =\;(3-2\sqrt{5}) - (3 + 2\sqrt{5}) \;=\;-4\sqrt{5}$

 

[pka]

Why did they write it that way? . . . Silly!

Therefore: .$\displaystyle \sqrt{29 - 12\sqrt{5}} - \sqrt{29+12\sqrt{5}} \:=\;\sqrt{(3-2\sqrt{5})^2} - \sqrt{(3 + 2\sqrt{5})^2}$
. . . . . . . . .$\displaystyle =\;(3-2\sqrt{5}) - (3 + 2\sqrt{5}) \;=\;-4\sqrt{5}$

I really do not understand the purpose of this mistaken reply.
$\displaystyle \sqrt{29 - 12\sqrt{5}} - \sqrt{29+12\sqrt{5}}=-6$

Here is the mistake:
$\displaystyle \sqrt{(3-2\sqrt{5})^2}\ne3-2\sqrt{5}$
Because $\displaystyle 3-2\sqrt{5}<0$ it must be $\displaystyle \sqrt{(3-2\sqrt{5})^2}=2\sqrt{5}-3$

I guess that is a Silly Mistake!