### 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 $$\blue 2x^2+x-3=0$$

Where we can take 1 as a common from the first two terms, and the equation becomes $$\blue x(2x+1)-3=0$$ 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. $$\blue x(2x-2)+x-3=0$$ $$\blue or \ x(2x-2)-1(2x-2)=0$$ $$\blue (2x-2)(x-1)=0$$ 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 $$3x^2-2x+5=0$$

As per hit-and-trial method, the values of a, b, and c are determined as $$a=3, \ b=-2, \ and \ c=5$$ So, we can write it as $$3x^2-3x+x+5=0$$ $$3x(x-1)+(x+5)=0$$ Taking 3x common from the first two terms, and 1 common from the last two terms, we get $$3x(x-1)+1(x+5)=0$$ $$(x+5)(3x-1)=0$$ $$\therefore x=\frac{-5}{3}, and \ \frac{1}{3}$$ Hence, the roots are -5/3 and 1/3 which are real and distinct.