MHB How to Solve Complex Algebraic Equations Involving Square Roots?

  • Thread starter Thread starter abrk
  • Start date Start date
AI Thread Summary
The discussion focuses on solving complex algebraic equations involving square roots. Two equations are presented, which require the correct interpretation of brackets, specifically that they represent (a + b√2)². Participants clarify the equations and derive a system of equations: a² + 2b² = 33 and ab = 20. For the third problem, the expression 11 - 6√2 is analyzed and shown to be a perfect square, leading to the conclusion that √(11 - 6√2) = 3 - √2. The thread emphasizes collaborative problem-solving and the importance of clear mathematical notation.
abrk
Messages
4
Reaction score
0
Hello.
I've been trying to crack these three for half a day and decided to ask for help.

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

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

3) Find: $$ \sqrt{11-6\sqrt{2}}$$.
 
Mathematics news on Phys.org
abrk said:
Hello.
I've been trying to crack these three for half a day and decided to ask for help.

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

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

3) Find: $$ \sqrt{11-6\sqrt{2}}$$.
You are missing a ")" in the first two problems. Since just [math](\sqrt{2})^2[/math] would make the problem trivial I assume you mean [math](a+ b\sqrt{2})^2[/math] in each.

Okay, in those two just go ahead and square on the left. I presume you know that [math](x+ y)^2= x^2+ 2xy+ y^2[/math].
[math](a+ b \sqrt{2})^2= a^2+ 2ab\sqrt{2}+ 2b^2[/math]
So in (1) you have [math]a^2+ 2b^2+ 2ab\sqrt{2}= 33+ 20\sqrt{2}[/math]. Assuming that a and b are supposed to be rational numbers, you must have [math]a^2+ 2b= 33[/math] and [math]2ab= 20[/math]

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!
 
HallsofIvy said:
[math]a^2+ 2b= 33[/math]
Hello,
You forgot $$a^2+2b^2=33$$, just wanted to tell that you forgot that so you can Edit! Have a nice day!
Regards,
$$|\pi\rangle$$
 
HallsofIvy said:
You are missing a ")" in the first two problems. Since just [math](\sqrt{2})^2[/math] would make the problem trivial I assume you mean [math](a+ b\sqrt{2})^2[/math] in each.

Okay, in those two just go ahead and square on the left. I presume you know that [math](x+ y)^2= x^2+ 2xy+ y^2[/math].
[math](a+ b \sqrt{2})^2= a^2+ 2ab\sqrt{2}+ 2b^2[/math]
So in (1) you have [math]a^2+ 2b^2+ 2ab\sqrt{2}= 33+ 20\sqrt{2}[/math]. Assuming that a and b are supposed to be rational numbers, you must have [math]a^2+ 2b= 33[/math] and [math]2ab= 20[/math]

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

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

Or in this case $$a^2+2b^2=33$$ and $$a*b=20$$

In 3, that's the whole description of the task, but a assume that it is to be represented as $$a+b\sqrt{2}$$, thus it is exactly the same as the first two problems.
 
Hello,
we got that:
(1)$$a^2+2b^2=33$$
(2)$$a*b=20$$
from (2) we get that $$a=\frac{20}{b}$$ and put that into (1) and solve it!
Good luck and have a nice day!
Regards,
$$|\pi\rangle$$
 
For the 3rd problem, I would observe that:

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

Now it should be straightforward to factor this as the square of a surd.
 
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}
 
I didn't even suspect that level of help and good will is possible nowadays.

Thank you very very much.
 
Back
Top