sir this is quadratic equation problem

Asked by Aravindh Nandha Raj s | 27th May, 2013, 06:04: PM

Expert Answer:

Let f(x) = (x-a)(x-c)+2(x-b)(x-d)
f(a) = 2(a-b)(a-d) > 0
f(b) = (b-a)(b-c) < 0
f(c) = 2(c-b)(c-d) < 0
f(d) = (d-a)(d-c) > 0

In going from a to b, f(x) goes from being greater than zero to being less than zero. Because f(x) is continuous, there is some a < x < b, such that f(x) = 0.
Similarly, there is some c < x < d, such that f(x) = 0. 

That is, there is one real root between a and b and another real root between c and d. Because f(x) is a polynomial of degree two, there are a total of two roots and they will be distinct

Answered by  | 28th May, 2013, 02:37: AM

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