# Nature Of Roots Of Quadratic Equation

Let us take the general form of a quadratic equation:

ax^{2}
+ 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 (b^{2} – 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 b

^{2}– 4ac > 0), then the roots and ß of equation (1) are

**real**and

**unequal.**

- 2. If discriminant is

**zero**(that is, if b

^{2}– 4ac = 0), then the roots and ß of equation (1) are

**real**and

**equal**.

- 3. If discriminant is

**negative**(that is, if b

^{2}– 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) ar

**real, irrational and unequal.**

- 5. If b

^{2}– 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 x^{2} – 18x + 81 = 0.**Solution:**

The discriminant of
the quadratic equation x^{2} – 18x + 81 = 0 is

(-18)^{2} – 4.1.81 = 324 – 324 = 0

Since the discriminant of the given equation is zero and coefficients of x^{2}, 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 ax^{2} + bx + c = 0 are rational.
**Solution:**

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

The given quadratic equation
is ax^{2} + bx + c = 0

Or ax^{2} –(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 px

^{2}– 2qx + p = 0 are real and unequal, then show that the roots of the equation qx

^{2}– 2px + q = 0 are imaginary.(Where both p and q are real)

**Solution:**

px

^{2}– 2qx + p = 0 (1)

qx

^{2}– 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(q

^{2}– p

^{2}) and

D2 = (-2p)

^{2}– 4.q.q = 4(

^{2}– q

^{2}).

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

D1 > 0 or 4(q

^{2}– p

^{2}) > 0 or q

^{2}> p

^{2}

Again q

^{2}> p

^{2}implies p

^{2}– q

^{2}< 0.

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

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