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Nature Of Roots Of Quadratic Equation

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Let us take the general form of a quadratic equation:

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

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

= dy/dx and = dy/dx

Now, suppose that a, b and c are real and rational. Then, the nature of the roots and 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 of equation (1):

    1. If discriminant is positive (that is, if b2 4ac > 0), then the roots and of equation (1) are real andunequal.
    2. If discriminant is zero (that is, if b2 4ac = 0), then the roots and of equation (1) are real and equal.
    3. If discriminant is negative (that is, if b2 4ac < 0), then the roots and 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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