If a+b+c=0 and a, b, c are rational, then the roots of the equation (b+c-a) x2+(c+a-b) x+(a+b-c)=0 are

Question

If a+b+c=0 and a, b, c are rational, then the roots of the equation (b+c-a) x2+(c+a-b) x+(a+b-c)=0 are (A) rational (B) irrational (C) imaginary (D) equal

Tardigrade Admin · Accepted Answer

We have, D=(c+a-b)2-4(b+c-a)(a+b-c) =(a+b+c-2 b)2-4(a+b+c-2 a) (a+b+c-2 c) =(-2 b)2-4(-2 a)(-2 c)=4(b2-4 a c) =4[(-a-c)2-4 a c]=4(a-c)2 =[2(a-c)]2= perfect square ∴ Roots are rational

Q. If $a + b + c = 0$ and $a, b, c$ are rational, then the roots of the equation $(b + c - a) x^{2} + (c + a - b) x + (a + b - c) = 0$ are

rational

irrational

imaginary

equal

We have,
$D = (c + a - b)^{2} - 4 (b + c - a) (a + b - c)$
$= (a + b + c - 2 b)^{2} - 4 (a + b + c - 2 a)$
$(a + b + c - 2 c)$
$= (- 2 b)^{2} - 4 (- 2 a) (- 2 c) = 4 (b^{2} - 4 a c)$
$= 4 [(- a - c)^{2} - 4 a c] = 4 (a - c)^{2}$
$= [2 (a - c)]^{2} =$ perfect square
$∴$ Roots are rational

Q. If a+b+c=0 and a,b,c are rational, then the roots of the equation (b+c−a)x2+(c+a−b)x+(a+b−c)=0 are