Question

Compare using the quadratic formula to find solutions to a quadratic equation having irrational roots to that of one that has rational roots.

Answer 1

For a general quadratic equation with rational coefficients

$ax^2+bx+c=0,\,\,\,\,a,b,c \in Q$

the two solutions are:

$x_{1,2} = \frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-b\pm\sqrt{D}}{2a}$

where D is the determinant.

Clearly, a solution will a rational number as long as $\sqrt{D}$ is rational. However, it can be shown that a square root of an integer is only rational if its value is an integer. In other words, $\sqrt{D}$ is rational if and only if the determinant is a perfect square, $D=n^2, \,\,\,n\in N$, otherwise the square root is irrational. Therefore the coefficients of quadratic equations that are to have rational solutions must satisfy the following condition:

$b^2-4ac=n^2\,\,\,n\in N$