What is the nature of the roots of a quadratic equation if the discriminant is negative?

• Math Doubts
• Quadratic equations
• Roots
• Nature

The roots of a quadratic equation are imaginary and distinct if the discriminant of a quadratic equation is negative.

Introduction

When a quadratic equation is expressed as $ax^2+bx+c = 0$ in algebraic form, the discriminant ($\Delta$ or $D$) of the quadratic equation is written as $b^2-4ac$.

The roots or zeros of the quadratic equation in terms of discriminant are written in the following two forms.

$(1).\,\,\,$ $\dfrac{-b+\sqrt{\Delta}}{2a}$

$(2).\,\,\,$ $\dfrac{-b-\sqrt{\Delta}}{2a}$

If the discriminant of the quadratic equation is negative, then the square root of the discriminant will be undefined. However, the square of a negative quantity can be expressed by an imaginary quantity.

For example $\sqrt{\Delta} \,=\, id$

Now, the zeros or roots of the quadratic equation can be written in the following form.

$(1).\,\,\,$ $\dfrac{-b+id}{2a}$

$(2).\,\,\,$ $\dfrac{-b-id}{2a}$

The two roots clearly reveal that the zeros or roots of the quadratic equation are distinct and imaginary.

Example

$5x^2+7x+6 = 0$

Evaluate the discriminant of this quadratic equation.

$\Delta \,=\, 7^2-4 \times 5 \times 6$

$\implies$ $\Delta \,=\, 49-120$

$\implies$ $\Delta \,=\, -71$

Similarly, find the square root of the discriminant.

$\implies$ $\sqrt{\Delta} \,=\, \sqrt{-71}$

$\implies$ $\sqrt{\Delta} \,=\, i\sqrt{71}$

The zeros or roots of the given quadratic equation are given here.

$\,\,\, \therefore \,\,\,\,\,\,$ $x \,=\, \dfrac{-7+i\sqrt{71}}{10}$ and $x \,=\, \dfrac{-7-i\sqrt{71}}{10}$

Therefore, it is proved that the roots are distinct and complex roots if the discriminant of quadratic equation is less than zero.

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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