# Surd

• Jan 25th 2008, 08:49 AM
slevvio
Surd
How do I simplify $\displaystyle \sqrt{9 + 4\sqrt{2}}$ to $\displaystyle 1+2\sqrt{2}$?

Thanks :)
• Jan 25th 2008, 10:13 AM
Soroban
Hello, slevvio!

Quote:

$\displaystyle \text{How do I simplify }\,\sqrt{9 + 4\sqrt{2}}\,\text{ to }\,1+2\sqrt{2}\;?$
I don't think there is an easy way . . .

First, we must suspect that $\displaystyle 9 + 4\sqrt{2}$ is a square.

Then we let: .$\displaystyle \sqrt{9 + 4\sqrt{2}} \;=\;a + b\sqrt{2}$
. . where both $\displaystyle a\text{ and }b$ are rational numbers.

Square both sides: .$\displaystyle 9 + 4\sqrt{2} \;=\;\left(a + b\sqrt{2}\right)^2$

. . and we have:. . $\displaystyle 9 + 4\sqrt{2} \;=\:(a^2 + 2b^2) + 2ab\sqrt{2}$

These two numbers are equal if their corresponding coefficients are equal.

So we have: .$\displaystyle \begin{Bmatrix}a^2+2b^2 & = & 9 \\ 2ab &=&4\end{Bmatrix}$

The second equation gives us: .$\displaystyle b \:=\:\frac{2}{a}$

Substitute into the first equation: .$\displaystyle a^2 + 2\left(\frac{2}{a}\right)^2 \:=\:9 \quad\Rightarrow\quad a^2 + \frac{8}{a^2} \:=\:9$

Multiply by $\displaystyle a^2\!:\;\;a^4 + 8 \:=\:9a^2\quad\Rightarrow\quad a^4 - 9a^2 + 8 \:=\:0$

. . which factors: .$\displaystyle (a^2-1)(a^2-8) \:=\:0$

. . and has roots: .$\displaystyle a \;=\;\pm1,\:\pm2\sqrt{2}$

Since $\displaystyle a$ must be rational and the original square root is positive,
. . the only acceptable root is: .$\displaystyle a\,=\,1\quad\Rightarrow\quad b \,=\,2$

Therefore: .$\displaystyle \sqrt{9 + 4\sqrt{2}} \;=\;1 + 2\sqrt{2}$

• Jan 25th 2008, 10:54 AM
Krizalid
Quote:

Originally Posted by Soroban
I don't think there is an easy way . . .

Of course there is one more dear Soroban :)

Quote:

Originally Posted by slevvio
How do I simplify $\displaystyle \sqrt{9 + 4\sqrt{2}}$ to $\displaystyle 1+2\sqrt{2}$?

$\displaystyle \sqrt {9 + 4\sqrt 2 } = \sqrt {\left( {1 + 4\sqrt 2 + 8} \right)} = \sqrt {\left( {1 + 2\sqrt 2 } \right)^2 } = \left| {1 + 2\sqrt 2 } \right|.$

Since the quantity inside bars is positive, we have that $\displaystyle \sqrt {9 + 4\sqrt 2 } = 1 + 2\sqrt 2 .$
• Jan 27th 2008, 08:49 AM
Soroban
Hello, Krizalid!

Of course, you are right . . . That is a valid method.

I use that method when I'm quite certain that we have a square.

I've done it enough times that I suspect that $\displaystyle 8 + 2\sqrt{15}$ is a square
. . because: .$\displaystyle 8 \:= \:3 + 5\;\hdots\;15 \:=\:3\cdot5\;\hdots\;\text{ and that coefficient is 2.}$

And, sure enough: .$\displaystyle 8 + 2\sqrt{15} \;=\;(\sqrt{3} + \sqrt{5})^2$