abstract algebra - $mathbb{Q}(sqrt p_{1}+sqrt p_{2})=mathbb{Q}(sqrt p_{1},sqrt p_{2})$ for $p_{1},p_{2}$ primes.

I want to prove that $\mathbb{Q}(\sqrt p_{1}+\sqrt p_{2})=\mathbb{Q}(\sqrt p_{1},\sqrt p_{2})$ for $p_{1},p_{2}$ primes. I know this was proved before for the generalized case of $p_{1},....,p_{n}$ primes here:

How to prove that $\mathbb{Q}[\sqrt{p_1}, \sqrt{p_2}, \ldots,\sqrt{p_n} ] = \mathbb{Q}[\sqrt{p_1}+ \sqrt{p_2}+\cdots + \sqrt{p_n}]$, for $p_i$ prime?

And every proof of this kind of equality involves Galois theory but I still don't know Galois theory and I'm supposed to prove this with only basic Field theory(field extensions, irreducible polynomials, algebraic extensions, etc).

By definition $\mathbb{Q}(\sqrt p_{1}+\sqrt p_{2})$ is the smallest field containing $\mathbb{Q}$ and $\sqrt p_{1}+\sqrt p_{2}$, also $\mathbb{Q}(\sqrt p_{1},\sqrt p_{2})$ is the smallest field containing $\sqrt p_{1}$ and $\sqrt p_{2}$. Or

$$\mathbb{Q}(\sqrt{p_{1}},\sqrt{p_{2}})=\{a+b\sqrt{p_{1}}+c\sqrt{p_{2}}+d\sqrt{p_{1}p_{2}} \mid a,b,c,d\in\mathbb{Q}\}$$

$$\mathbb{Q}(\sqrt{p_{1}}+\sqrt{p_{2}}) = \lbrace a+b(\sqrt{p_{1}}+\sqrt{p_{2}}) \mid a,b \in \mathbb{Q} \rbrace$$

Also I know that $[\mathbb{Q}(\sqrt p_{1},\sqrt p_{2}):\mathbb{Q}]=4$ from this:

Proving that $\left(\mathbb Q[\sqrt p_1,\dots,\sqrt p_n]:\mathbb Q\right)=2^n$ for distinct primes $p_i$.

Still don't know how to proceed proving this two extensions are the same.

Answer

Hint

Step 1 :

$$\sqrt{p_1}-\sqrt{p_2}=\frac{p_1-p_2}{\sqrt{p_1}+\sqrt p_2}\in \mathbb Q(\sqrt{p_1}+\sqrt{p_2})$$

Step 2 :

$$\sqrt{p_1}=\frac{(\sqrt{p_1}+\sqrt p_2)+(\sqrt{p_1}-\sqrt{p_2})}{2}\in \mathbb Q(\sqrt{p_1}+\sqrt{p_2}).$$

I let you manage to show that $\sqrt{p_2}\in \mathbb Q(\sqrt{p_1}+\sqrt{p_2})$ as well and conclude the equality.