SOLUTION: Given that x and y are integers and {{{matrix(1,3,sqrt(8sqrt(11)+27),""="",y-sqrt(x-2y))}}}, , find {{{-y^x}}}

Algebra ->  Radicals -> SOLUTION: Given that x and y are integers and {{{matrix(1,3,sqrt(8sqrt(11)+27),""="",y-sqrt(x-2y))}}}, , find {{{-y^x}}}      Log On


   



Question 1151181: Given that x and y are integers and
matrix%281%2C3%2Csqrt%288sqrt%2811%29%2B27%29%2C%22%22=%22%22%2Cy-sqrt%28x-2y%29%29, , find -y%5Ex

Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!
Your equation is 

matrix%281%2C3%2Csqrt%288sqrt%2811%29%2B27%29%2C%22%22=%22%22%2Cy-sqrt%28x-2y%29%29, find -y%5Ex

After working on it, I found that there is no solution in integers.

However if the minus sign before the square root on the right side is changed to
plus, there is a solution.  I will prove that I am right. 

I'll first do it with the minus sign, just as was given, show that there is no
solution, then change that minus sign to plus, and find the solution.

First we simplify the left side by writing it as the sum of two square roots of
integers.

sqrt%28A%29%2Bsqrt%28B%29, where A and B are positive integers.

matrix%281%2C3%2Csqrt%288sqrt%2811%29%2B27%29%2C%22%22=%22%22%2Csqrt%28A%29%2Bsqrt%28B%29%29

Square both sides:

(1)    matrix%281%2C3%2C8sqrt%2811%29%2B27%2C%22%22=%22%22%2CA%2B2sqrt%28A%29sqrt%28B%29%2BB%29

Equate the terms with square roots:

matrix%281%2C3%2C8sqrt%2811%29%2C%22%22=%22%22%2C2sqrt%28A%29sqrt%28B%29%29 

Divide both sides by 2

matrix%281%2C3%2C4sqrt%2811%29%2C%22%22=%22%22%2Csqrt%28A%29sqrt%28B%29%29

Square both sides

matrix%281%2C3%2C%2816%29%2811%29%2C%22%22=%22%22%2C%28A%29%28B%29%29

matrix%281%2C3%2C176%2C%22%22=%22%22%2CAB%29

Equate the terms of (1) that do not have square roots:

matrix%281%2C3%2C27%2C%22%22=%22%22%2CA%2BB%29

So we have the system of equations:

system%28AB=176%2CA%2BB=27%29

Solve that by substitution and get {A,B} = {11,16}

So the left side of the original equation is

sqrt%2811%29%2Bsqrt%2816%29

or

sqrt%2811%29%2B4

We might stop here to convince ourselves with a calculator
that the left side of the original equation actually does
equal to the above expression.

matrix%281%2C3%2Csqrt%288sqrt%2811%29%2B27%29%2C%22%22=%22%22%2Csqrt%2811%29%2B4%29

By calculator, we get the same approximation 7.31662479 for both expressions,
so that satisfies us that the left side of the original equation does simplify
to sqrt%2811%29%2B4

Now the original equation becomes:

matrix%281%2C3%2Csqrt%2811%29%2B4%2C%22%22=%22%22%2Cy-sqrt%28x-2y%29%29

matrix%281%2C3%2Csqrt%2811%29%2Bsqrt%28x-2y%29%2C%22%22=%22%22%2Cy-4%29

Since y is an integer, so is y-4.  

Suppose that integer is A.  Then

matrix%281%2C3%2Csqrt%2811%29%2Bsqrt%28x-2y%29%2C%22%22=%22%22%2CA%29
matrix%281%2C3%2Csqrt%28x-2y%29%2C%22%22=%22%22%2CA-sqrt%2811%29%29
Square both sides:
matrix%281%2C3%2Cx-2y%2C%22%22=%22%22%2CA%5E2-2Asqrt%2811%29%2B11%29
matrix%281%2C3%2C2Asqrt%2811%29%2C%22%22=%22%22%2CA%5E2%2B11-x-2y%29

Since x, y, and A are all integers the right side is also an integer.
However the left side is not an integer but is irrational.

Therefore I have proved that your equation as stated has no solution.
That's because an integer cannot equal to an irrational number.

However if the original equation had been this:

matrix%281%2C3%2Csqrt%288sqrt%2811%29%2B27%29%2C%22%22=%22%22%2Cy%2Bsqrt%28x-2y%29%29

Then we would have had this:

matrix%281%2C3%2Csqrt%2811%29%2B4%2C%22%22=%22%22%2Cy%2Bsqrt%28x-2y%29%29

The two sides will be exactly the same if y=4 and x-2y=11

Substituting 4 for y in 

x-2y=11
x-2(4)=11
x-8=11
x=19

So the solution with the minus changed to plus is x=19, y=4.

Then, -y%5Ex=-19%5E4=-130321

You should call your teacher's attention to this.  It has to be an error.

Edwin