Shortcut Methods
JEE Mains
Coefficient of x^2 = 2 Coefficient of x = 1 Constant term = -3 Discriminant D=25>0
This represents a quadratic equation such as
Where we can take 1 as a common from the first two terms, and the equation becomes
Further, we will split the -3 into -2 and -1 or -3 and -1, the product of which will be -3 and their sum will be 1, the coefficient of x. Hence the roots are x = 2/1 and 1. Both roots are real and distinct.
CBSE Class 11 and Class 12 Exams
Coefficient of x^2 = 3 Coefficient of x = -2 Constant term = 5 Discriminant D>0 (Since roots are real)
The quadratic equation is
As per hit-and-trial method, the values of a, b, and c are determined as
So, we can write it as Taking 3x common from the first two terms, and 1 common from the last two terms, we get Hence, the roots are -5/3 and 1/3 which are real and distinct.
