First slide

Theory of equations

Question

If $a, b, c$ are real, then both the roots of the equation
$(x - b) (x - c) + (x - c) (x - a) + (x - a) (x - b) = 0$ (1)
are always

Moderate

A

positive
B

negative
C

real
D

none of these

Solution

We can write (1) as
$3 x^{2} - 2 (a + b + c) x + b c + c a + a b = 0$ (2)
The discriminant D of (2) is given by
$\begin{array}{l} D = 4 (a + b + c)^{2} - 4 \times 3 (b c + c a + a b) \\ = 4 [a^{2} + b^{2} + c^{2} + 2 b c + 2 c a + 2 a b - 3 b c - 3 c a - 3 a b] \\ = 4 [a^{2} + b^{2} + c^{2} - b c - c a - a b] \\ = 2 [(b^{2} + c^{2} - 2 b c) + (c^{2} + a^{2} - 2 c a) + (a^{2} + b^{2} - 2 a b)] \\ = 2 [(b - c)^{2} + (c - a)^{2} + (a - b)^{2}] \end{array}$
As $a, b, c$ are real, we get $D \geq 0 .$ Thus, roots of (1) are real.

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