# Nature of the Roots of a Quadratic Equation

We will discuss here about the different cases of discriminant to understand the nature of the roots of a quadratic equation.

We know that α and β are the roots of the general form of the quadratic equation ax$$^{2}$$ + bx + c = 0 (a ≠ 0) .................... (i) then we get

α = $$\frac{- b - \sqrt{b^{2} - 4ac}}{2a}$$ and β = $$\frac{- b + \sqrt{b^{2} - 4ac}}{2a}$$

Here a, b and c are real and rational.

Then, the nature of the roots α and β of equation ax$$^{2}$$ + bx + c = 0 depends on the quantity or expression i.e., (b$$^{2}$$ - 4ac) under the square root sign.

Thus the expression (b$$^{2}$$ - 4ac) is called the discriminant of the quadratic equation ax$$^{2}$$ + bx + c = 0.

Generally we denote discriminant of the quadratic equation by ‘∆ ‘ or ‘D’.

Therefore,

Discriminant ∆ = b$$^{2}$$ - 4ac

Depending on the discriminant we shall discuss the following cases about the nature of roots α and β of the quadratic equation ax$$^{2}$$ + bx + c = 0.

When a, b and c are real numbers, a ≠ 0

Case I: b$$^{2}$$ - 4ac > 0

When a, b and c are real numbers, a ≠ 0 and discriminant is positive (i.e., b$$^{2}$$ - 4ac > 0), then the roots α and β of the quadratic equation ax$$^{2}$$ + bx + c = 0 are real and unequal.

Case II: b$$^{2}$$ - 4ac = 0

When a, b and c are real numbers, a ≠ 0 and discriminant is zero (i.e., b$$^{2}$$ - 4ac = 0), then the roots α and β of the quadratic equation ax$$^{2}$$ + bx + c = 0 are real and equal.

Case III: b$$^{2}$$ - 4ac < 0

When a, b and c are real numbers, a ≠ 0 and discriminant is negative (i.e., b$$^{2}$$ - 4ac < 0), then the roots α and β of the quadratic equation ax$$^{2}$$ + bx + c = 0 are unequal and imaginary. Here the roots α and β are a pair of the complex conjugates.

Case IV: b$$^{2}$$ - 4ac > 0 and perfect square

When a, b and c are real numbers, a ≠ 0 and discriminant is positive and perfect square, then the roots α and β of the quadratic equation ax$$^{2}$$ + bx + c = 0 are real, rational unequal.

Case V: b$$^{2}$$ - 4ac > 0 and not perfect square

When a, b and c are real numbers, a ≠ 0 and discriminant is positive but not a perfect square then the roots of the quadratic equation ax$$^{2}$$ + bx + c = 0 are real, irrational and unequal.

Here the roots α and β form a pair of irrational conjugates.

Case VI: b$$^{2}$$ - 4ac is perfect square and a or b is irrational

When a, b and c are real numbers, a ≠ 0 and the discriminant is a perfect square but any one of a or b is irrational then the roots of the quadratic equation ax$$^{2}$$ + bx + c = 0 are irrational.

Notes:

(i) From Case I and Case II we conclude that the roots of the quadratic equation ax$$^{2}$$ + bx + c = 0 are real when b$$^{2}$$ - 4ac ≥ 0 or b$$^{2}$$ - 4ac ≮ 0.

(ii) From Case I, Case IV and Case V we conclude that the quadratic equation with real coefficient cannot have one real and one imaginary roots; either both the roots are real when b$$^{2}$$ - 4ac > 0 or both the roots are imaginary when b$$^{2}$$ - 4ac < 0.

(iii) From Case IV and Case V we conclude that the quadratic equation with rational coefficient cannot have only one rational and only one irrational roots; either both the roots are rational when b$$^{2}$$ - 4ac is a perfect square or both the roots are irrational b$$^{2}$$ - 4ac is not a perfect square.

Various types of Solved examples on nature of the roots of a quadratic equation:

1. Find the nature of the roots of the equation 3x$$^{2}$$ - 10x + 3 = 0 without actually solving them.

Solution:

Here the coefficients are rational.

The discriminant D of the given equation is

D = b$$^{2}$$ - 4ac

= (-10)$$^{2}$$ - 4  3  3

= 100 - 36

= 64 > 0.

Clearly, the discriminant of the given quadratic equation is positive and a perfect square.

Therefore, the roots of the given quadratic equation are real, rational and unequal.

2. Discuss the nature of the roots of the quadratic equation 2x$$^{2}$$ - 8x + 3 = 0.

Solution:

Here the coefficients are rational.

The discriminant D of the given equation is

D = b$$^{2}$$ - 4ac

= (-8)$$^{2}$$ - 4  2 ∙ 3

= 64 - 24

= 40 > 0.

Clearly, the discriminant of the given quadratic equation is positive but not a perfect square.

Therefore, the roots of the given quadratic equation are real, irrational and unequal.

3. Find the nature of the roots of the equation x$$^{2}$$ - 18x + 81 = 0 without actually solving them.

Solution:

Here the coefficients are rational.

The discriminant D of the given equation is

D = b$$^{2}$$ - 4ac

= (-18)$$^{2}$$ - 4  1  81

= 324 - 324

= 0.

Clearly, the discriminant of the given quadratic equation is zero and coefficient of x$$^{2}$$ and x are rational.

Therefore, the roots of the given quadratic equation are real, rational and equal.

4. Discuss the nature of the roots of the quadratic equation x$$^{2}$$ + x + 1 = 0.

Solution:

Here the coefficients are rational.

The discriminant D of the given equation is

D = b$$^{2}$$ - 4ac

= 1$$^{2}$$ - 4  1  1

= 1 - 4

= -3 > 0.

Clearly, the discriminant of the given quadratic equation is negative.

Therefore, the roots of the given quadratic equation are imaginary and unequal.

Or,

The roots of the given equation are a pair of complex conjugates.

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