Question

If


`√(x^2+2y+4) +√(x^2+x-y+5)=√(x^2+x+3y+2) +√(x^2+2x+3)`

`x+y=?`

Answer 1

x + y = 2

Explanation:

You want to solve an equation in x and y that is the sum of two square roots equal to a different sum of two square roots.

Solution

This necessarily gets messy. You must square the equation, rearrange it to isolate the radicals, and repeat that process two more times. A number of terms cancel, so the result is a 4th degree equation in x and y that can be factored to the square of the product of two linear trinomials.

Here goes:

`2x^2 +x +y +9 +2√((5 + x + x^2 - y)(4 + x^2 + 2 y))\\\\=2 x^2 +3x + 3 y+5 + 2√((3 + 2 x + x^2)(2 + x + x^2 + 3 y))\qquad\text{square both sides}\\\\\\x+y-2=√((5 + x + x^2 - y)(4 + x^2 + 2 y))-√((3 + 2 x + x^2)(2 + x + x^2 + 3 y))\\\text{ separate radicals from non-radical terms}\\\\`

Repeating the process of squaring both sides and separating the radical terms, we get ...

`22 + 15 x + 15 x^2 + 4 x^3 + 2 x^4 + 19 y + 6 x y + 4 x^2 y - 3 y^2\\\\=2√((3 + 2 x + x^2)(5 + x + x^2 - y)(4 + x^2 + 2 y)(2 + x + x^2 + 3 y))`

Ater one final squaring of both sides, we get an 8th-degree equation that can be factored with some difficulty to ...

`(2 + x - x^2 - 7 y + 2 x y + 3 y^2)^2 = 0\\\\(x -3y +1)^2(x+y-2)^2=0`

Zero product rule

Solutions to this are values of x and y that make the factors zero. The second factor is zero when ...

(x +y -2) = 0 ⇒ x +y = 2 . . . . . . . . the value the problem is looking for

The first factor is zero when ...

x -3y +1 = 0 ⇒ x = 3y -1

Using this expression to substitute for x in the original radical equation, we get ...

`2√(5 - 4 y + 9 y^2)=2√(2 + 9 y^2)\\\\-4y+3=0\qquad\text{divide by 2, square, subtract $2+9y^2$}\\\\y=(3)/(4)\\\\x=3\cdot(3)/(4)-1=(5)/(4)\\\\x+y=(5)/(4)+(3)/(4)=2`

The solution to this equation is x +y = 2.

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Additional comment

As we did with x=3y-1, we can substitute x=2-y into the original radical equation. Doing so gives a tautology, good for all values of x and y.

Attached is a graph of the radical equation. It is the line x+y=2.