Solve for x & y: $a+b\sqrt{2}$ Given $2x+y-\sqrt{3x^2+3xy+y^2}=2+\sqrt{2}$

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In summary, the given equation has two variables and only one equation, so the values of x and y cannot be determined. However, it can be solved algebraically by isolating the square root terms and squaring both sides of the equation. This equation has infinite solutions for x and y, and the square root terms make it more challenging to solve algebraically. Other methods such as graphical and numerical methods can also be used to solve this equation.
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Given $x$ and $y$ are of the form $a+b\sqrt{2}$ ($a$ and $b$ are both positive integers) that satisfy the equation

$2x+y-\sqrt{3x^2+3xy+y^2}=2+\sqrt{2}$.

Find such $x$ and $y$.
 
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anemone said:
Given $x$ and $y$ are of the form $a+b\sqrt{2}$ ($a$ and $b$ are both positive integers) that satisfy the equation

$2x+y-\sqrt{3x^2+3xy+y^2}=2+\sqrt{2}$.

Find such $x$ and $y$.

Rearranging and squaring, then rearranging again (with the quadratic formula in mind), we arrive at

\(\displaystyle x=\dfrac{4(2+\sqrt2)-y+\sqrt{y^2+12(2+\sqrt2)^2}}{2}=\dfrac{4(2+\sqrt2)-y+\sqrt{y^2+72+48\sqrt2}}{2}\)

We now seek positive integers $a,b,c,d$ such that

\(\displaystyle a^2+2b^2+72=c^2+2d^2\)

and

\(\displaystyle 2ab+48=2cd\).

By inspection, $a=b=1$ and $c=d=5$, so $x=6+4\sqrt2,y=1+\sqrt2$ are one such $x$ and $y$.
 
  • #3
greg1313 said:
Rearranging and squaring, then rearranging again (with the quadratic formula in mind), we arrive at

\(\displaystyle x=\dfrac{4(2+\sqrt2)-y+\sqrt{y^2+12(2+\sqrt2)^2}}{2}=\dfrac{4(2+\sqrt2)-y+\sqrt{y^2+72+48\sqrt2}}{2}\)

We now seek positive integers $a,b,c,d$ such that

\(\displaystyle a^2+2b^2+72=c^2+2d^2\)

and

\(\displaystyle 2ab+48=2cd\).

By inspection, $a=b=1$ and $c=d=5$, so $x=6+4\sqrt2,y=1+\sqrt2$ are one such $x$ and $y$.

Thanks greg1313 for participating! Your solution is neat and elegant, good job!

My solution (which is clearly more convoluted than yours:eek:):

Let $x=a+b\sqrt{2},\,y=c+d\sqrt{2}$, where $a,\,b,\,c,\,d\in \Bbb{N}$.

Squaring the given equality, and rearrange it, we have:

$x^2+(y-8-4\sqrt{2})x+(6+4\sqrt{2}-4y-2\sqrt{2}y)=0$

$x^2+(c-8+(d-4)\sqrt{2})x+(6-4c-4d+(4-4d-2c)\sqrt{2})=0$

Assume the quadratic in $x$ has roots $x_1=a+b\sqrt{2}$ and $x_2=a-b\sqrt{2}$, we find $d=4,\,c=-6$ but we're told $c>0$, so our assumption is wrong.

Next, it's safe to assume that the quadratic in $x$ above has roots $x_1=a+2\sqrt{2}$ and $x_2=p-2\sqrt{2}$, where $p$ is an integer, and $b=2$, and comes from

$\underline{(x-(a+2\sqrt{2}))(x-(p-2\sqrt{2}))}(x-(a-2\sqrt{2}))(x-(p+2\sqrt{2}))=0$

Again, we find $d=4$. We find $ap=-2(1+2c),\,a+p=8-c,\,p-a=-6-c$. If we let $p=-2$, then we get $c=3$ and therefore $a=7$.

Thus, one of the solutions for $(x,\,y)$ is $(7+2\sqrt{2},\,3+4\sqrt{2})$.
 

FAQ: Solve for x & y: $a+b\sqrt{2}$ Given $2x+y-\sqrt{3x^2+3xy+y^2}=2+\sqrt{2}$

What is the value of x and y in the given equation?

The values of x and y cannot be determined from the given equation as there are two variables and only one equation. We need at least two equations to solve for both x and y.

Can this equation be solved algebraically?

Yes, this equation can be solved algebraically by isolating the square root terms and squaring both sides of the equation to eliminate the radical.

Is there more than one solution for x and y?

Yes, there are infinite solutions for x and y in this equation. This is because there are two variables and only one equation, so any values that satisfy the equation can be considered as solutions.

What is the importance of the square root terms in this equation?

The square root terms are important because they introduce non-linear terms in the equation, making it more challenging to solve algebraically.

Can this equation be solved using other methods besides algebra?

Yes, this equation can also be solved using graphical methods, such as graphing both sides of the equation and finding the points of intersection. It can also be solved using numerical methods, such as substitution or elimination.

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