Nature of Roots of a Quadratic Equation

have seen that the roots of the equation ax² + bx + c = 0 are given by

`x=(-b+-sqrt(b^2-4ac))/(2a)`

If b² – 4ac > 0, we get two distinct real roots, `-b/(2a)+(b^2-4ac)/(2a)` and `-b/(2a)-(sqrt(b^2-4ac))/(2a)`

If b² – 4ac = 0,then x= `-b/(2a)+-0` i.e., `x=-b/(2a) or -b/(2a)`

So, the roots of the equation ax² + bx + c = 0 are both `-b/(2a)`

Therefore, we say that the quadratic equation ax² + bx + c = 0 has two equal real roots in this case.

If b² – 4ac < 0, then there is no real number whose square is b² – 4ac. Therefore, there are no real roots for the given quadratic equation in this case.

Since b² – 4ac determines whether the quadratic equation ax² + bx + c = 0 has real roots or not, b2 – 4ac is called the discriminant of this quadratic equation.
So, if a quadratic equation ax² + bx + c = 0 has

(i) two distinct real roots, if b² – 4ac > 0,
(ii) two equal real roots, if b² – 4ac = 0,
(iii) no real roots, if b² – 4ac < 0.
Let us consider one examples.

Example  : Find the discriminant of the quadratic equation 2x² – 4x + 3 = 0, and hence find the nature of its roots.
Solution : The given equation is of the form ax² + bx + c = 0, where a = 2, b = – 4 and c = 3.

Therefore, the discriminant
b² – 4ac = (– 4)² – (4 × 2 × 3) = 16 – 24 = – 8 < 0
So, the given equation has no real roots.