Question

Both the roots of the equation, (if given that a,b,c are real) (x-b)(x-c) + (x-a)(x-c) + (x-a)(x-b) = 0 are always :

• A. Positive
• B. Negative
• C. Real
• D. None of these

Answer 1

The correct option is C Real

Solve using assumption

Let a = 0, b = 1, c = 2. The expression becomes 3x² -6x + 2 = 0. The discriminant of this exp is positive. You can try with some other values. You will always get real roots.

Alternatively: -

(x-b)(x-c) + (x-a)(x-c) + (x-a)(x-b) = 0

x² - cx - bx + bc + x² - cx - ax + ac + x² - bx - ax + ab = 0

3x² - 2(a + b + c)x + (ab + bc + ca) = 0

Discriminant = b² - 4ac

4(a + b + c)² - 4 x 3 x (ab + bc + ca)

4{a² + b² + c² + 2(ab + bc + ca) - 3(ab + bc + ca)}

4{a² + b² + c² - (ab + bc + ca)}

The discriminant of this exp is positive. You will always get real roots.