Hints:
1) Define
$\phi:\Bbb Q[x]\to\Bbb Q[\sqrt d]\,\,\,,\,\,\phi(f(x)):=f(\sqrt d)$
and show the above is a ring homomorphism, find its kernel and use the first isomorphism theorem for rings.
Further hint for this one: You may want to show that for $\,g(x)\in\Bbb [x]\,$ ,
$g(\sqrt d)=0\Longleftrightarrow (x^2-d)\mid g(x)$
2) Prove that a ring homomorphism as wanted sends a root of some irreducible polynomial to another root of the same irred. polynomial.
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General hint Using the same notation as in the question, $\,\Bbb Q[\sqrt d]=\Bbb Q(\sqrt d)\,$ , i.e.: this ring is actually a field.
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3) Prove that if $\,F_p:=\Bbb Z/p\Bbb Z\,=$ the prime with $\,p \,$elements , $\,p\,$ a prime, then
$f(x)\in\Bbb F_p[x]\;\;\text{is irreducible}\,\,\,\Longleftrightarrow \,\,\text{the ideal}\,\,(f(x))\in\Bbb F_p[x]\,\,\,\text{is maximal}\,$
$\Longleftrightarrow \,\Bbb F_p[x]/(f(x))\,\,\,\text{is a field}$
and show that in this last case, the field has $\,p^{\deg f}\,$ elements . Use then that there exists one single field, up to isomorphism, of any finite order (which, of course, is always a power of a prime)