Question

If $a,b,c$ are distinct rational numbers, then the roots of the quadratic equation $(a+b-2c)x^2+(b+c-2a)x+(c+a-2b)=0$ are

• A. rational and distinct
• B. irrational and distinct
• C. irrational and equal
• D. rational and equal

Answer 1

The correct option is A rational and distinct

$(a+b-2c)x^2+(b+c-2a)x+(c+a-2b)=0\Rightarrow a(x^2-2x+1)+b(x^2+x-2)-c(2x^2-x-1)=0\Rightarrow a(x-1{)}^2+b[(x-1)(x+2)]-c[(2x+1)(x-1)]=0\Rightarrow (x-1\left)\right[a(x-1)+b(x+2)-c(2x+1)]=0\Rightarrow (x-1\left)\right[x(a+b-2c)-(a+c-2b)]=0\Rightarrow x=1,\frac{a+c-2b}{a+b-2c}$
$\therefore$ Roots are rational and distinct.