FIND THE NATURE OF ROOTS OF THE QUADRATIC EQUATION

Nature of the roots of the Equation ax2 + bx + c = 0

The two roots of the Quadratic equation ax2 + bx + c = 0 are:

x = -b±b2 - 4ac2a

The expression b2 – 4ac which appear under radical sign is called the Discriminant (Disc.) of the quadratic equation. i.e., Disc = b2 – 4ac

The expression b2 – 4ac discriminates the nature of the roots, whether they are real, rational, irrational or imaginary. There are three possibilities.

Condition

b2 – 4ac > 0

Nature

is a perfect square, the roots are real, rational and unequal.

is not a perfect square, then roots are real, irrational and unequal.

 b2 – 4ac = 0

then roots will be real, equal and rational.

 b2 – 4ac < 0

then roots will be imaginary and unequal.

Example 1 :

Find the nature of the roots of the following equations

2x2 + 3x + 1 = 0

Solution :

Comparing the given quadratic equation with ax2 + bx + c = 0, we get

a = 2, b = 3 and c = 1

b2 - 4ac = 32 - 4(2) (1)

= 9-8

= 1 > 0

Since it is perfect square, roots are real, rational and unequal.

Example 2 :

Find the nature of the roots of the following equations

6x2 = 7x + 5

Solution :

6x2 = 7x + 5

Converting into standard form, we get

6x2 - 7x - 5 = 0

Comparing the given quadratic equation with ax2 + bx + c = 0, we get

a = 6, b = -7 and c = -5

b2 - 4ac = (-7)2 - 4(6) (-5)

= (49 + 120)

= 169 > 0

Since it is perfect square, roots are real, rational and unequal.

Example 3 :

Find the nature of the roots of the following equations

3x2 + 7x - 2 = 0

Solution :

3x2 + 7x - 2 = 0

Comparing the given quadratic equation with ax2 + bx + c = 0, we get

a = 3, b = 7 and c = -2

b2 - 4ac = 72 - 4(3) (-2)

= (49 + 24)

= 73 > 0

Since it is not perfect square, roots are real, irrational and unequal.

Example 4 :

Find the nature of the roots of the following equations

√2x2 + 3x - √8 = 0

Solution :

√2x2 + 3x - √8 = 0

Comparing the given quadratic equation with ax2 + bx + c = 0, we get

a = √2, b = 3 and c = -√8

b2 - 4ac = 32 - 4(√2) (-√8)

= 9 + 16

= 25 > 0

Since it is perfect square, roots are real, rational and unequal.

Example 5 :

Show that the roots of the following equations are rational

a(b – c)x2 + b(c – a)x + c(a – b) = 0

Solution :

a = a(b - c), b = b(c - a) and c = c(a - b)

b2 - 4ac = [b(c - a)]2 - 4a(b - c) c(a - b)

= b2(c - a)2 - 4ac(b - c)(a - b)

= b2(c2 - 2ac + a2) - 4ac(ab - b2 - ac + bc)

=  b2 c2 - 2acb2 + a2b2 - 4a2bc + 4acb2 + 4a2c2 - 4abc2

= a2b+ 4a2c2 b2 c2 - 4a2bc - 4abc+2 acb2

= (ab)2 + (-2ac)2 + (bc)2 - 4a2bc - 4abc+2 acb2

= (ab - 2ac + bc)2

Since it is perfect square, the given quadratic equation will have rational roots.

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