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Question: The roots of the equation a(b –2c)x<sup>2</sup> + b(c –2a) x + c (a –2b) = 0 are, when ab + bc + ca ...

The roots of the equation a(b –2c)x2 + b(c –2a) x + c (a –2b) = 0 are, when ab + bc + ca = 0

A

1, c(a−2b)a(b−2c)\frac{c(a - 2b)}{a(b - 2c)}

B

ca\frac{c}{a}, a−2bb−2c\frac{a - 2b}{b - 2c}

C

a−2ba−2c\frac{a - 2b}{a - 2c}, a−2bb−2c\frac{a - 2b}{b - 2c}

D

None of these

Answer

1, c(a−2b)a(b−2c)\frac{c(a - 2b)}{a(b - 2c)}

Explanation

Solution

As it is given that ab + bc + ca = 0 so putting

x = 1 in the equation we get

f(1) = a(b –2c) + b(c –2a) + c(a – 2b)

⇒ f(1) = –Σab = 0

so 1 is a root of the equation

Now product of the roots be

1. α = c(a−2b)a(b−2c)\frac{c(a - 2b)}{a(b - 2c)}

⇒ α = ca\frac { \mathrm { c } } { \mathrm { a } } (a−2bb−2c)\left( \frac{a - 2b}{b - 2c} \right)