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

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.

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

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.

Latest Math Topics

Latest Math Problems

Email subscription

Math Doubts is a best place to learn mathematics and from basics to advanced scientific level for students, teachers and researchers.
Know more

Learn how to solve easy to difficult mathematics problems of all topics in various methods with step by step process and also maths questions for practising.