Square root problem.

[kreshnik]

Hello.

\(\displaystyle \sqrt{11-6\sqrt{2}}+\sqrt{11+6\sqrt{2}}=?\)

\(\displaystyle A)\;4\)

\(\displaystyle B)\;6\)

\(\displaystyle C)\;8\)

\(\displaystyle D)\;10\)

Attempted:

\(\displaystyle \sqrt{11-6\sqrt{2}}+\sqrt{11+6\sqrt{2}}= \text{Stuck!}\)

 

[Deleted member 4993]

Hello.

\(\displaystyle \sqrt{11-6\sqrt{2}}+\sqrt{11+6\sqrt{2}}=?\)

\(\displaystyle A)\;4\)

\(\displaystyle B)\;6\)

\(\displaystyle C)\;8\)

\(\displaystyle D)\;10\)

Attempted:

\(\displaystyle \sqrt{11-6\sqrt{2}}+\sqrt{11+6\sqrt{2}}= \text{Stuck!}\)

Hint:

\(\displaystyle 11 - 6\sqrt{2} \ = \ 3^2 + (\sqrt{2})^2 - 2 * 3 * \sqrt{2}\)

 

[kreshnik]

Ohh.. I'm sorry, my bad, I should have known that:

\(\displaystyle \sqrt{\frac{\sqrt[3]{3^9}-26}{1}+2+\sqrt5\cdot\sqrt5+8-5-\sqrt{\frac{144}{4}}\cdot\sqrt{2}}=\sqrt{11-6\sqrt2}\)

You call that a hint??...

 

[lookagain]

Hello.

\(\displaystyle \sqrt{11-6\sqrt{2}}+\sqrt{11+6\sqrt{2}}=?\)

\(\displaystyle A)\;4\)

\(\displaystyle B)\;6\)

\(\displaystyle C)\;8\)

\(\displaystyle D)\;10\)

Attempted:

\(\displaystyle \sqrt{11-6\sqrt{2}}+\sqrt{11+6\sqrt{2}}= \text{Stuck!}\)

kreshnik,

please look at the methods above, and do those if possible.

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Note: The use of the word "between" below is not inclusive for the
numbers at either end.


However, only because there are multiple choices and the gap between
each adjacent pairs is 2, then consider this estimation method:



Rewrite as \(\displaystyle \sqrt{11 - \sqrt{72}} \ + \ \sqrt{11 + \sqrt{72}} = \)


\(\displaystyle \sqrt{11 - (\# \ between \ 8 \ and \ 9)} \ + \ \sqrt{11 + (\# \ between \ 8 \ and \ 9)} \ = \)


\(\displaystyle \sqrt{\# \ between \ 2 \ and \ 3} \ + \ \sqrt{\# \ between \ 19 \ and \ 20} \ = \)


\(\displaystyle (\# \ between \ 1 \ and \ 2) \ + \ (\# \ between \ 4 \ and \ 5) \ = \)


\(\displaystyle \# \ between \ 5 \ and \ 7\)


Of the offered values, the only choice that fits that is 6.


Therefore, the answer is B).

 

[soroban]

Hello, kreshnik!

\(\displaystyle \text{Evaluate: }\:\sqrt{11-6\sqrt{2}}+\sqrt{11+6\sqrt{2}}\)

. . \(\displaystyle (A)\;4 \qquad (B)\;6 \qquad (C)\;8 \qquad (D)\;10\)

When I see the form: .\(\displaystyle \sqrt{p + q\sqrt{r}}\)

. . I suspect that \(\displaystyle p + q\sqrt{r}\) is a square.


I conjecture that: .\(\displaystyle a + b\sqrt{2} \:=\:\sqrt{11+6\sqrt{2}}\)

Square both sides: .\(\displaystyle \left(a + b\sqrt{2}\right)^2 \:=\:\left(\sqrt{11+6\sqrt{2}}\right)^2\)

. . . . . . . . . \(\displaystyle a^2 + 2ab\sqrt{2} + 2b^2 \:=\:11+6\sqrt{2}\)


Equate rational and irrational components:

. . . . . \(\displaystyle \begin{Bmatrix}a^2 + 2b^2 &=& 11 & [1] \\ 2ab\sqrt{2} &=& 6\sqrt{2} & [2]\end{Bmatrix}\)


From [2]: .\(\displaystyle b \:=\:\dfrac{3}{a}\;\;[3]\)

Substitute into [1]: .\(\displaystyle a^2 + 2\left(\dfrac{3}{a}\right)^2 \:=\:11 \quad\Rightarrow\quad a^2 + \dfrac{18}{a^2} \:=\:11 \quad\Rightarrow\quad a^4 + 18 \:=\:11a^2\)

. . . . . . . . . . . . . . .\(\displaystyle a^4 - 11a^2 + 18 \:=\:0 \quad\Rightarrow\quad (a^2 - 9)(a^2 - 2) \:=\:0 \)

. . . . . . . . . . . . . . .\(\displaystyle \begin{Bmatrix}a^2-9\:=\:0 & \Rightarrow & a \;=\;\pm3 \\ a^2-2 \:=\:0 & \Rightarrow & a =\pm\sqrt{2} \end{Bmatrix}\)

Substitute into [3]: .\(\displaystyle \begin{Bmatrix} b \;=\;\pm1 \\ b =\pm\frac{3}{\sqrt{2}} \end{Bmatrix} \)


And we find that: .\(\displaystyle \begin{Bmatrix}11 + 6\sqrt{2} \;=\;\big[\pm(3 + \sqrt{2})\big]^2 & \Rightarrow & \sqrt{11 + 6\sqrt{2}} \;=\;3 + \sqrt{2} \\ 11 - 6\sqrt{2} \:=\:\big[\pm(3 - \sqrt{2})\big]^2 & \Rightarrow & \sqrt{11-6\sqrt{2}} \;=\;3 - \sqrt{2} \end{Bmatrix}\)