MHB For which integers x,y is (4-6*sqrt(2))^2 = x+y*sqrt(2)?

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The equation (4-6*sqrt(2))^2 can be expressed in the form x + y*sqrt(2) by expanding it correctly. The first method attempted led to incorrect values, but the correct expansion yields 88 - 48*sqrt(2). Thus, the integers x and y are 88 and -48, respectively. This demonstrates the importance of accurately expanding and simplifying expressions involving square roots. The solution highlights the necessity of careful algebraic manipulation in such problems.
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Hi All

I have the following question.View attachment 5866

I have reviewed my notes but have not been able to crack this.

I tried two different ways, both wrong.

First:

$$
(4-6*\sqrt2)^2=$$

$$16-24*\sqrt2-24*\sqrt2+(36*2)

= 88-218*\sqrt2$$

so, $x=88$ and

$y=218$My second method was

$$(4-6*\sqrt2)^2= 4^2-(6*2)=28$$

but this is not in the form they want.

I'd really appreciate some advice on how to go about solving this kind of problem.

Thanks!

Daniel
 

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I think your first method is the way I would use, so that the LHS of the equation is in the form $a+b\sqrt{2}$:

$$(4-6\sqrt{2})^2=x+y\sqrt{2}$$

$$16-48\sqrt{2}+72=x+y\sqrt{2}$$

$$88-48\sqrt{2}=x+y\sqrt{2}$$

And so we see that:

$$(x,y)=(88,-48)$$
 
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