Question

If $a,b,c\in Q$, then roots of the equation $(b+c-2a)x^2+(c+a-2b)x+(a+b-2c)=0$ are:

• A. irrational
• B. rational
• C. none of these
• D. non real

Answer 1

The correct option is B rational

For quadratic equation $a{x}^2+bx+c=0,a\ne 0$
Nature of roots is given by discriminant $D=b^2-4ac$ as
$\left(i\right)D>0\Rightarrow \text{Real and Distinct roots}\left(ii\right)D=0\Rightarrow \text{Real and equal roots}\left(iii\right)D<0\Rightarrow \text{Non real or imaginary roots}$
Also the roots of the above equation are given by
$\frac{-b\pm \sqrt{D}}{2a}$

Given equation is $(b+c-2a)x^2+(c+a-2b)x+(a+b-2c)=0$
$D=(c+a-2b{)}^2-4(b+c-2a\left)\right(a+b-2c)$
$=\left(\right(c+a{)}^2+4b^2-4(c+a)b-4\left(\right(b+c\left)\right(a+b)-2c(b+c)-2a(a+b)+4ac))$
$=c^2+a^2+2ac+4b^2-4bc-4ab-4(ab+b^2+ac+bc-2bc-2c^2-2a^2-2ab+4ac)$
$=c^2+a^2+2ac+4b^2-4bc-4ab-4ab-4b^2-4ac-4bc+8bc+8c^2+8a^2+8ab-16ac$
$=9a^2+9c^2-18ac$
$=(3a-3c{)}^2$
$=(3a-3c{)}^2$
which is a perfect square and is greater than or equal to zero
$\Rightarrow D\ge 0$ and a perfect square.
So the roots will be rational.