Question

Prove that the roots of the equation $(a-b+c)x^2+2(a-b)x+(a-b-c)=0$ are rational numbers for all real numbers a and b and for all rational c.

Answer 1

Let the given equation be of the form $A{x}^2+Bx+C=0$.

Then, $A=a-b+c,B=2(a-b)$ and $C=a-b-c$.

Now, the discriminant of $A{x}^2+Bx+C=0$ is

$B^2-4AC=\left[2\right(a-b){]}^2-4(a-b+c\left)\right(a-b-c)$

$=4(a-b{)}^2-4[(a-b)+c\left]\right[(a-b)-c]$

$=4(a-b{)}^2-4[(a-b{)}^2-c^2]$

$\Delta =4(a-b{)}^2-4(a-b{)}^2+4c^2=4c^2$, a perfect square.

Therefore, $\Delta >0$ and it is a perfect square.

Hence, the roots of the given equation are rational numbers.