Can the roots of a quadratic equation ever be a square root plus another square root?

Question

Can one of the roots of a quadratic equation ever be of the form $\sqrt{x} + \sqrt{y}$? Assuming $x$ and $y$ are not perfect squares.

The coefficients and constant of the quadratic equation need to be integers. I'm looking to see if $\sqrt{x} + \sqrt{y}$ can ever be an algebraic number of degree 2.

Answer 1

It's very easy to construct. $$ p(z) = (z-(\sqrt{x} + \sqrt{y}))(z-a) $$ is such a polynomial for any $a$. You can put any $x$ and $y$ you like in there.

Perhaps you want to ask the question with more restrictions.

Answer 2

Sure. Just work backwards and force it.

$x = \sqrt{3} + \sqrt{11}$

$x^2 = 3 + 2\sqrt{33} + 11 = 14 +2\sqrt{33}$

$x^2 -14 = 2\sqrt{33}$

$(x^2 - 14)^2 = 4*33$

$x^4 - 28x^2 + 14^2 = 132$

$x^4 - 28x^2 + 64=0$ has $\sqrt{3} + \sqrt{11}$ as one of its roots. ... which one wouldn't think by looking at it.

.....

Or in general $x = \sqrt{x} + \sqrt{y}$

$w^2 = x + y + 2\sqrt{xy}$

$(w^2 - (x+y))^2 = 4xy$

$w^4 -2(x+y)w^2 + (x^2 - 2xy + y^2)=0$

$w^4 - 2(x+y)w^2 + (x-y)^2= 0$.

Hey, that's kind of nicely symmetric.