Question

If $a+b+c=0$, prove that the roots of $a{x}^2+bx+c=0$ are rational. Hence, show that the roots of $(p+q)x^2-2px+(p-q)=0$ are rational.

Answer 1

$a+b+c=0$

$b=-a-c$

$\Rightarrow b=-(a+c)$

given by $a{x}^2+bx+c$

$D=b^2-4ac$

$=(-(a+c){)}^2-4ac$ ,(putting the value of b)

$=(a+c{)}^2-4ac=a^2+c^2+2ac-4ac$

$=a^2+c^2-2ac$

$=(a-c{)}^2$

thus, the root

$\frac{-b\pm \sqrt{D}}{2a}$

$=\frac{-b\pm \sqrt{(a-c{)}^2}}{2a}$

$=\frac{-b+a-c}{2a}$

thus, we can conclude that if discriminant is a perfect square then root are rational

$(p+q)x^2-2px+(p-q)=0$

$D=(-2p{)}^2-4(p+q\left)\right(p-q)$

$=4p^2-4(p^2-q^2)$ $\left(as\right(a+b\left)\right(a-b)=(a^2-b^2\left)\right)$

$=4p^2-4p^2+4q^2=4q^2$

$=(2q{)}^2$

hence, the root must be rational.