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:

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.

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

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

= (ab)² + (-2ac)² + (bc)² - 4a²bc - 4abc² +2 acb²

= (ab - 2ac + bc)²

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