Question

What is the value of x if (a + 2b – 3c)x^2 + (b + 2c – 3a)x + (c + 2a – 3b) = 0 where a, b, c are in A.P?

(a) 1/2

(b) 1/4

(c) 2/3

(d) 3/4

The question was posed to me in a job interview.

My question comes from Applications of Quadratic Equations topic in division Complex Numbers and Quadratic Equations of Mathematics – Class 11

Answer 1

Right choice is (b) 1/4

Easiest explanation: If the coefficients of (x^2 + x + c) = 0, then

x will always be = 1

Therefore, here, (a + 2b – 3c) + (b + 2c – 3a) + (c + 2a – 3b) = 0

So, x = 1.

As, one of its root is 1 so, we will calculate the other one.

As, a, b, c are in A.P so,

b = (a + c)/2

Thus, product of the roots αβ = (c + 2a – 3b)/(a + 2b – 3c)

As, a root say α = 1, then,

β = (c + 2a – 3(a + c)/2) / (a + 2(a + c)/2 – 3c)

We get the value of β = 1/4