If a, b, c are real, then both the roots of the equation (x-b)(x-c)+(x-c)(x-a)+(x-a)(x-b)=0 are always

Question

If a, b, c are real, then both the roots of the equation (x-b)(x-c)+(x-c)(x-a)+(x-a)(x-b)=0 are always (A) positive (B) negative (C) real (D) imaginary

Tardigrade Admin · Accepted Answer

Given equation can be rewritten as 3x2 - 2x(a + b + c)+ ab+ bc + ca = 0 Now, Discriminant, D = 4(a + b + c)2 - 4⋅3(ab + bc + ca) =4 (a2+b2+c2-ab-bc-ca) =2[(a-b)2+(b-c)2+(c-a)2] ge 0 Hence, roots are real

Q. If $a, b, c$ are real, then both the roots of the equation $(x - b) (x - c) + (x - c) (x - a) + (x - a) (x - b) = 0$ are always

positive

negative

real

imaginary

Given equation can be rewritten as

$3 x^{2} - 2 x (a + b + c) + ab + b c + c a = 0$

Now, Discriminant,

$D = 4 (a + b + c)^{2} - 4 \cdot 3 (ab + b c + c a)$

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

$= 2 [(a - b)^{2} + (b - c)^{2} + (c - a)^{2}] \geq 0$

Hence, roots are real

Q. If a,b,c are real, then both the roots of the equation (x−b)(x−c)+(x−c)(x−a)+(x−a)(x−b)=0 are always