Hint $\ $ If a field F has two F-linear independent combinations of $\rm\ \sqrt{a},\ \sqrt{b}\ $ then you can solve for $\rm\ \sqrt{a},\ \sqrt{b}\ $ in F. For example, the Primitive Element Theorem works that way, obtaining two such independent combinations by Pigeonholing the infinite set $\rm\ F(\sqrt{a} + r\ \sqrt{b}),\ r \in F,\ |F| = \infty,$ into the finitely many fields between F and $\rm\ F(\sqrt{a}, \sqrt{b}),$ e.g. see PlanetMath's proof.
In this case it's simpler to notice $\rm\ E = \mathbb Q(\sqrt{a} + \sqrt{b})\ $ contains the independent $\rm\ \sqrt{a} - \sqrt{b}\ $ since
$\rm \sqrt{a}\ -\, \sqrt{b}\,\ =\ \dfrac{\ a\:-\:b}{\sqrt{a}\,+\sqrt{b}}\ \in\ E = \mathbb Q(\sqrt{a}+\sqrt{b}) $
To be explicit, notice that $\rm\ u = \sqrt{a}+\sqrt{b},\,\ v = \sqrt{a}-\sqrt{b}\in E\ $ so solving the linear system for the roots yields $\rm\ \sqrt{a}\ =\ (u+v)/2,\ \ \sqrt{b}\ =\ (v-u)/2\:,\ $ both of which are clearly $\rm\:\in E\:,\:$ since $\rm\:u,\:v\in E\:$ and $\rm\:2\ne 0\:$ in $\rm\:E\:,\:$ so $\rm\:1/2\:\in E\:.\:$ This works over any field where $\rm\:2\ne 0\:,\:$ i.e. where the determinant (here $2$) of the linear system is invertible, i.e. where the linear combinations $\rm\:u,v\:$ of the square-roots are linearly independent over the base field.