Surds - Problem

abrk

New member
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: $$\displaystyle \sqrt{11-6\sqrt{2}}$$.

HallsofIvy

Well-known member
MHB Math Helper
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: $$\displaystyle \sqrt{11-6\sqrt{2}}$$.
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.

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 mean by "find" that number. The way it is given is perfectly valid. I presume you do NOT mean just use a calculator!

Petrus

Well-known member
$$\displaystyle a^2+ 2b= 33$$
Hello,
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,
$$\displaystyle |\pi\rangle$$

abrk

New member
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.

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 mean by "find" that number. The way it is given is perfectly valid. I presume you do NOT mean just use a calculator!
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 system

$$\displaystyle a^2+b^2=c$$
$$\displaystyle a*b=d$$

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.

Petrus

Well-known member
Hello,
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,
$$\displaystyle |\pi\rangle$$

MarkFL

Staff member
For the 3rd problem, I would observe that:

$$\displaystyle 11-6\sqrt{2}=3^2-2\cdot3\sqrt{2}+\sqrt{2}^2$$

Now it should be straightforward to factor this as the square of a surd.

soroban

Well-known member
Hello, abrk!

HallsofIvy gave you hints for the first two problems.

$$\text{3) Find: }\;\sqrt{11-6\sqrt{2}}$$

By now you may suspect that $$11-6\sqrt{2}$$ is a perfect square.

Let $$(a + b\sqrt{2})^2 \:=\:11-6\sqrt{2}$$
. . where $$a > 0$$ and $$a,b$$ are rational numbers.

Then: .$$a^2 + 2b^2 + 2\sqrt{2}ab \:=\:11-6\sqrt{2}$$

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

From [2]: .$$b \,=\,\text{-}\tfrac{3}{a}\;\;[3]$$
Substitute into [1]: .$$a^2 + 2\left(\text{-}\tfrac{3}{a}\right)^2 \:=\:11$$

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

. . $$a \;=\;\pm\sqrt{2},\;\pm3$$

Hence: .$$\boxed{a \,=\,3}$$

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

Hence: .$$(3-\sqrt{2})^2 \:=\:11 - 6\sqrt{2}$$

Therefore: .$$\sqrt{11-6\sqrt{2}} \;=\;3-\sqrt{2}$$

abrk

New member
I didn't even suspect that level of help and good will is possible nowadays.

Thank you very very much.