Quadratic EquationMedium
Question
Both the roots of the equation (x - b)(x - c) + (x - a)(x - c) + (x - a)(x - b) = 0 are always
Options
A.positive
B.negative
C.real
D.None of these
Solution
(x - a)(x - b) + (x - b)(x - c) + (x - c)(x - a) = 0
⇒ 3x2 - 2(a + b + c)x + (ab + bc + ca) = 0
Now, discriminant
= 4(a + b + c)2 - 12(ab + bc + ca)
= 4{a2 + b2 + c2 - ab - bc - ca}
= 2{(a - b)2 + (b - c)2 + (c - a)2}
which is always positive.
Hence, both roots are real.
⇒ 3x2 - 2(a + b + c)x + (ab + bc + ca) = 0
Now, discriminant
= 4(a + b + c)2 - 12(ab + bc + ca)
= 4{a2 + b2 + c2 - ab - bc - ca}
= 2{(a - b)2 + (b - c)2 + (c - a)2}
which is always positive.
Hence, both roots are real.
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