- Thread starter
- #1
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: \(\displaystyle \sqrt{11-6\sqrt{2}}\).
Hello,\(\displaystyle a^2+ 2b= 33\)
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 you mean by "find" that number. The way it is given is perfectly valid. I presume you do NOT mean just use a calculator!
[tex]\text{3) Find: }\;\sqrt{11-6\sqrt{2}}[/tex]