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Simplifying roots of surds

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  • #1

MarkFL

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Feb 24, 2012
13,775
Without the use of a calculator, and showing your work, simplify:

\(\displaystyle \frac{1}{2}\left(\left(239+169\sqrt{2}\right)^{ \frac{1}{7}}-\left(29\sqrt{2}-41\right)^{ \frac{1}{5}}\right)\)

edit: My apologies...I was careless in my first statement of the problem...(Nod)
 
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soroban

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Feb 2, 2012
409
Hello, MarkFL!

Without the use of a calculator, and showing your work, simplify:

. . [tex]\tfrac{1}{2}\left(\left(239+169\sqrt{2}\right)^{ \frac{1}{7}}-\left(29\sqrt{2}-41\right)^{\frac{1}{5}}\right)[/tex]

We find that: .[tex]239 + 169\sqrt{2} \:=\: (1+\sqrt{2})^7[/tex]

. . .and that: .. [tex]29\sqrt{2} - 41 \:=\: (\sqrt{2} - 1)^5[/tex]

So we have: .[tex]\tfrac{1}{2}\left(\left[(1+\sqrt{2})^7\right]^{\frac{1}{7}} - \left[(\sqrt{2}-1)^5\right]^{\frac{1}{5}}\right)[/tex]

. . . . . . . [tex]=\;\tfrac{1}{2}\bigg(\left[1 + \sqrt{2}\right] - \left[\sqrt{2}-1\right]\bigg)[/tex]

. . . . . . . [tex]=\;\tfrac{1}{2}\big(1 + \sqrt{2} - \sqrt{2} + 1\big)[/tex]

. . . . . . . [tex]=\;\tfrac{1}{2}(2)[/tex]

. . . . . . . [tex]=\;1[/tex]
 

topsquark

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MHB Math Helper
Aug 30, 2012
1,123
@Soroban. You are the master of coming up with solutions that seem to come out of the blue. How the heck did you get the first two lines of your solution???

-Dan
 

Bacterius

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MHB Math Helper
Jan 26, 2012
644
@Soroban. You are the master of coming up with solutions that seem to come out of the blue. How the heck did you get the first two lines of your solution???

-Dan
My thoughts exactly. I was sitting there, thinking, "okay". You must teach me! (Rofl)
 

kaliprasad

Well-known member
Mar 31, 2013
1,309
(239+169√2)^1/7

the root can be of the form (a+b √2) and expanding and equating the rational parts of both sides and irrational parts of both sides we get 2 eqautions in a and b
then solving them we get a = b = 1

I know these are polynomials of degree 7 and 6 and solving is not simple so assuming a and b as integers
we can put the values on both sides and get a and b.

simlilarly for we can find 5th root of 2nd expression
 
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MarkFL

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Feb 24, 2012
13,775
This is the observation I made before constructing this problem:

\(\displaystyle \left(\sqrt{2}+1 \right)^0=0\cdot\sqrt{2}+1\)

\(\displaystyle \left(\sqrt{2}+1 \right)^1=1\cdot\sqrt{2}+1\)

\(\displaystyle \left(\sqrt{2}+1 \right)^2=2\cdot\sqrt{2}+3\)

\(\displaystyle \left(\sqrt{2}+1 \right)^3=5\cdot\sqrt{2}+7\)

\(\displaystyle \left(\sqrt{2}+1 \right)^4=12\cdot\sqrt{2}+17\)

Now, we may generalize to say:

\(\displaystyle \left(\sqrt{2}+1 \right)^n=U_n\sqrt{2}+V_n\)

We may further generalize and write:

\(\displaystyle \left(\sqrt{2}-1 \right)^n=(-1)^{n-1}\left(U_n\sqrt{2}-V_n \right)\)

The parameters may be defined recursively as:

\(\displaystyle U_{n+1}=2U_{n}+U_{n-1}\) where \(\displaystyle U_0=0,\,U_1=1\)

\(\displaystyle V_{n+1}=2V_{n}+V_{n-1}\) where \(\displaystyle V_0=1,\,V_1=1\)

We then find:

\(\displaystyle U_5=29,\,U_7=169\)

\(\displaystyle V_5=41,\,V_7=239\)

and so:

\(\displaystyle 29\sqrt{2}-41=(-1)^4\left(\sqrt{2}-1 \right)^5\,\therefore\,\left(29\sqrt{2}-41 \right)^{\frac{1}{5}}=\sqrt{2}-1\)

\(\displaystyle 169\sqrt{2}+239=\left(\sqrt{2}+1 \right)^7\,\therefore\,\left(169\sqrt{2}+239 \right)^{\frac{1}{7}}=\sqrt{2}+1\)

Hence:

\(\displaystyle \frac{1}{2}\left(\left(169\sqrt{2}+239 \right)^{\frac{1}{7}}-\left(29\sqrt{2}-41 \right)^{\frac{1}{5}} \right)=\frac{1}{2}\left(\sqrt{2}+1-\sqrt{2}+1 \right)=\frac{1}{2}\cdot2=1\)