PROBLEM: Given $f:$ and $g:$ are two quadratic polynomials with rational coefficients.
Suppose $f(x)=0$ and $g(x)=0$ have a common irrational solution.
Prove that $f(x)=rg(x)$ for all $x$ where $r$ is a rational number.

SOLUTION: Suppose the common irrational root of (\ f(x)) and (\ g(x)) be (\sqrt{a}+b).

Then by properties of irrational roots we can say that the other root of both of them will be (\sqrt{a}-b).

so we can write (\ f(x)=\lambda(x-\sqrt{a}-b)(x-\sqrt{a}+b)) and (\ g(x)=\mu(x-\sqrt{a}-b)(x-\sqrt{a}+b))

so (\frac{g(x)}{f(x)}=\frac{\mu}{\lambda})

therefore,$$\ g(x)=f(x)\frac{\mu}{\lambda}=rf(x)$$.


Theorem:In an equation with real coefficients irrational roots occurs in conjugate pairs.