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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