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.

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