Be rational

Algebra Level 3

$f(x)$ is a quadratic polynomial in $x$ with rational coefficients.

Given $f(\sqrt{2})=3$ and $f(\sqrt{3})=2$ , what is $\lfloor 100 f(\sqrt{5}) \rfloor$ ?

1 solution

Pulkit Gupta
Dec 21, 2015

Lets define the quadratic as $\large ax^{2} + bx + c$ .

Then f ( $\sqrt{2}$ ) = $\large 2a + \sqrt{2} b + c$ = 3 ; f ( $\sqrt{3}$ ) = $\large 3a + \sqrt{3} b + c$ = 2

The coefficients are rational so for the R.H.S to be an integer, the coefficient with the irrational term must be equal to zero. Therefore, b = ZERO.

On solving the equations obtained on putting b = 0, we get a = -1 and c =5.

By the above, f ( $\sqrt{5}$ ) = $\large 5(-1) + 5$ = 0