# Simplifying roots of surds

#### MarkFL

##### Administrator
Staff member
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...

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#### soroban

##### Well-known member
Hello, MarkFL!

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

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

We find that: .$$239 + 169\sqrt{2} \:=\: (1+\sqrt{2})^7$$

. . .and that: .. $$29\sqrt{2} - 41 \:=\: (\sqrt{2} - 1)^5$$

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

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

. . . . . . . $$=\;\tfrac{1}{2}\big(1 + \sqrt{2} - \sqrt{2} + 1\big)$$

. . . . . . . $$=\;\tfrac{1}{2}(2)$$

. . . . . . . $$=\;1$$

#### topsquark

##### Well-known member
MHB Math Helper
@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

##### Well-known member
MHB Math Helper
@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!

#### kaliprasad

##### Well-known member
(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

#### MarkFL

##### Administrator
Staff member
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:

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