The quadratic equations \( 2006 x^{2}+2007 x+1=0 \) \( \& x^{2}+2007 x+2006=0 \)

Question

4. The quadratic equations \( 2006 x^{2}+2007 x+1=0 \) \( \& x^{2}+2007 x+2006=0 \) have a common root. Then product of the uncommon roots is (A) -1 (B) 1 (C) 2006 (D) 2007

Answer 1

rambha6810, hello

The roots of the first equation are -1/2006 and -1

The roots of the second equation are -1 and -2006

Then the product of the uncommon roots is (B) 1 

Answer 2

let a be common root then

2006a² + 2007a + 1  = 0   ..... Eqn 1

a² + 2007a + 2006  = 0 ..... Eqn 2

Subtract Eqn 2 from Eqn 1

2005a² - 2005 = 0 then  root a = -1, +1

value of a which satisfies Enq 1 and Eqn 2   is -1

let p be another root of Eqn 1

Sum of roots = a + p = -2007/2006
p = -2007/2006 + 1  = -1/2006

let q be another root of Eqn 2

Sum of roots = a + q = -2007
q = -2007 + 1  = - 2006

Product of other roots pq = (-1/2006)(- 2006)  = 1