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