Question

Assertion (A): The roots of $(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0$ are real
Reason (R): A quadratic equation with non-negative discriminant has real roots .

• A. Both (A) and (R) are true and (R) is the correct explanation of (A)
• B. Both (A) and (R) are true and (R) is not the correct explanation of (A)
• C. (A) is true but (R) is false
• D. (A) is false but (R) is true

Answer 1

Answer: (A)

Both (A) and (R) are true and (R) is the correct explanation of (A)

Roots of quadratic equation are real if its discriminant $b^2-4ac\ge 0$ .
Given equation can be written in expanded form as,

$x^2-ax-bx+ab+x^2-bx-cx+bc+x^2-cx-ax+ac=0$
$3x^2-2(a+b+c)x+ab+bc+ca=0$

Let $\Delta$ be the discriminant of the above quadratic equation
$\Delta =4(a+b+c{)}^2-4(3\left)\right(ab+bc+ca)$
$=4(a^2+b^2+c^2-ab-bc-ca)$
$=4\left(\frac{1}{2}\right(a-b{)}^2+(b-c{)}^2+(c-a{)}^2)$
$=2\left(\right(a-b{)}^2+(b-c{)}^2+(c-a{)}^2)$

Thus,
$\Delta \ge 0$ always .