Let us take the general form of a quadratic equation:

ax² + bx + c = 0     [a ≠ 0 ]                      (1)

If and ß be the roots of the equation (1), then

= and ß =

Now, suppose that a, b and c are real and rational. Then, the nature of the roots and ß of
equation (1) is determinedby the expression (b2 – 4ac) under the radical sign.

Therefore (b² – 4ac) is known as the Discriminant of equation (1). Referred to this discriminant
following conclusions can be drawn about the nature of roots and ß of equation (1):

1. If discriminant is positive (that is, if b² – 4ac > 0), then the roots and ß of equation (1) are real andunequal.

2. If discriminant is zero (that is, if b² – 4ac = 0), then the roots and ß of equation (1) are real and equal.

3. If discriminant is negative (that is, if b² – 4ac < 0), then the roots and ß of equation (1) are imaginary and unequal.

4. If discriminant is positive and a perfect square then the roots of equation (1) are real, rational and unequal. And if the discriminant is positive but not a perfect square then the roots of equation (1) arreal, irrational and unequal.

5. If b² – 4ac is a perfect square but any one of a or b is irrational then the roots of equation (1) are irrational.

Example 1:

Discuss the nature of the roots of the equation x² – 18x + 81 = 0.

Solution:

The discriminant of the quadratic equation x² – 18x + 81 = 0 is

(-18)² – 4.1.81 = 324 – 324 = 0

Since the discriminant of the given equation is zero and coefficients of x², x are rational, hence roots of the equation are real, rational and equal.

Example 2:

If a, b, c are rational and a + b + c = 0, show that the roots of the equation ax² + bx + c = 0 are rational.

Solution:

Since a + b + c = 0 hence b = -(c + a)

The given quadratic equation is ax² + bx + c = 0

Or ax² –(c + a)x + c = 0 (1)

The discriminant of equation (1) is {-(c + a)}² – 4ac = (c + a)² – 4ac = (c - a) ²

Since a, b, c are rational and the discriminant of equation (1) is a perfect square, hence the roots of (1) are rational.

Example 3

If the roots of the equation px² – 2qx + p = 0 are real and unequal, then show that the roots of the equation qx² – 2px + q = 0 are imaginary.(Where both p and q are real)

Solution:

px² – 2qx + p = 0             (1)

qx² – 2px + q = 0             (2)

Let D1 and D2 be the discriminants of equations (1) and (2) respectively.

Then D1 = (-2q)² – 4.p.p = 4(q² – p²) and

D2 = (-2p) ² – 4.q.q = 4(² – q²).

Given that p and q are real and the roots of equation (1) are real and unequal, hence we get,

D1 > 0 or 4(q² – p²) > 0 or q² > p²

Again q² > p² implies p² – q² < 0.

Hence the discriminant of equation (2) is negative. For this reason, the roots of the equation (2) are imaginary.

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