# Nature of Roots of a Quadratic Equation

#### notes

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.

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Quadratic Equation part 10 (Quadratic Formula, Nature of roots) [00:12:47]
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