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- Thread starter abrk
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- Jan 29, 2012

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You are missing a ")" in the first two problems. Since just \(\displaystyle (\sqrt{2})^2\) would make the problem trivial I assume you mean \(\displaystyle (a+ b\sqrt{2})^2\) in each.Hello.

I've been trying to crack these three for half a day and decided to ask for help.

1) \(\displaystyle (a+b(\sqrt{2})^2=33+20(\sqrt{2})\)

2) \(\displaystyle (a+b(\sqrt{2})^2=41-24(\sqrt{2})\)

3) Find: \(\displaystyle \sqrt{11-6\sqrt{2}}\).

Okay, in those two just go ahead and square on the left. I presume you know that \(\displaystyle (x+ y)^2= x^2+ 2xy+ y^2\).

\(\displaystyle (a+ b \sqrt{2})^2= a^2+ 2ab\sqrt{2}+ 2b^2\)

So in (1) you have \(\displaystyle a^2+ 2b^2+ 2ab\sqrt{2}= 33+ 20\sqrt{2}\). Assuming that a and b are supposed to be rational numbers, you must have \(\displaystyle a^2+ 2b= 33\) and \(\displaystyle 2ab= 20\)

For (3), I don't know what you

- Feb 21, 2013

- 739

Hello,\(\displaystyle a^2+ 2b= 33\)

You forgot \(\displaystyle a^2+2b^2=33\), just wanted to tell that you forgot that so you can Edit! Have a nice day!

Regards,

\(\displaystyle |\pi\rangle\)

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For 1 and 2. Indeed, there was one more bracket, which I didn't embed in my original question. My whole problem is that I cannot get through the equation systemYou are missing a ")" in the first two problems. Since just \(\displaystyle (\sqrt{2})^2\) would make the problem trivial I assume you mean \(\displaystyle (a+ b\sqrt{2})^2\) in each.

Okay, in those two just go ahead and square on the left. I presume you know that \(\displaystyle (x+ y)^2= x^2+ 2xy+ y^2\).

\(\displaystyle (a+ b \sqrt{2})^2= a^2+ 2ab\sqrt{2}+ 2b^2\)

So in (1) you have \(\displaystyle a^2+ 2b^2+ 2ab\sqrt{2}= 33+ 20\sqrt{2}\). Assuming that a and b are supposed to be rational numbers, you must have \(\displaystyle a^2+ 2b= 33\) and \(\displaystyle 2ab= 20\)

For (3), I don't know what youmeanby "find" that number. The way it is given is perfectly valid. I presume you do NOT mean just use a calculator!

\(\displaystyle a^2+b^2=c\)

\(\displaystyle a*b=d\)

Or in this case \(\displaystyle a^2+2b^2=33\) and \(\displaystyle a*b=20\)

In 3, that's the whole description of the task, but a assume that it is to be represented as \(\displaystyle a+b\sqrt{2}\), thus it is exactly the same as the first two problems.

- Feb 21, 2013

- 739

we got that:

(1)\(\displaystyle a^2+2b^2=33\)

(2)\(\displaystyle a*b=20\)

from (2) we get that \(\displaystyle a=\frac{20}{b}\) and put that in to (1) and solve it!

Good luck and have a nice day!

Regards,

\(\displaystyle |\pi\rangle\)

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- #6

HallsofIvy gave you hints for the first two problems.

[tex]\text{3) Find: }\;\sqrt{11-6\sqrt{2}}[/tex]

By now you may suspect that [tex]11-6\sqrt{2}[/tex] is a perfect square.

Let [tex](a + b\sqrt{2})^2 \:=\:11-6\sqrt{2}[/tex]

. . where [tex]a > 0[/tex] and [tex]a,b[/tex] are rational numbers.

Then: .[tex]a^2 + 2b^2 + 2\sqrt{2}ab \:=\:11-6\sqrt{2}[/tex]

Equate coefficients: .\begin{Bmatrix}a^2+2b^2 \:=\:11 & [1] \\ 2ab \:=\:\text{-}6 & [2] \end{Bmatrix}

From [2]: .[tex]b \,=\,\text{-}\tfrac{3}{a}\;\;[3][/tex]

Substitute into [1]: .[tex]a^2 + 2\left(\text{-}\tfrac{3}{a}\right)^2 \:=\:11 [/tex]

[tex]a^4 - 11a^2 + 18 \:=\:0 \quad\Rightarrow\quad (a^2-2)(a^2-9) \:=\:0[/tex]

. . [tex]a \;=\;\pm\sqrt{2},\;\pm3[/tex]

Hence: .[tex]\boxed{a \,=\,3}[/tex]

Substitute into [3]: .[tex]b \,=\,\text{-}\tfrac{3}{3} \quad\Rightarrow\quad \boxed{b \,=\,\text{-}1}[/tex]

Hence: .[tex](3-\sqrt{2})^2 \:=\:11 - 6\sqrt{2}[/tex]

Therefore: .[tex]\sqrt{11-6\sqrt{2}} \;=\;3-\sqrt{2}[/tex]

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