The roots of the quadratic equation ( a + b -2c )x² – ( 2a – b – c)x +( a – 2b + c) = 0 are-
A. (a + b + c ) and (a – b- c)
B. \( \frac{1}{2} \) and a – 2b + c
C. a – 2b + c and \( \frac{1}{(a + b – 2c)} \)
D. None of the above.
Answer
Clearly we see that x=1 satisfies the equation
⇒x=1 is a root of the that equation
Let α be other root
Product of roots =1 × α = \( \frac{a – 2b + c}{a + b – 2c} \)
other root = α = \( \frac{a – 2b + c}{a + b – 2c} \)
how do we state the nature of quadratics
A polynomial can have real numbers as zero(ie rational and irrational)to decide it’s nature we can use the relationship between the coefficients and zero (by comparing sum and product of zeroes)