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Irrational root(Tomato subjective 28)

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.

September 19, 2016

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