The roots of a quadratic equation are not real but they are imaginary complex conjugates and it is possible only when the discriminate of a quadratic equation is less than zero.

The discriminant ($\Delta$) of a quadratic equation $ax^2+bx+c = 0$ is $b^2-4ac$.

The roots of a quadratic equation in terms of discriminant are $\dfrac{-b + \sqrt{\Delta}}{2a}$ and $\dfrac{-b \,-\sqrt{\Delta}}{2a}$

If the discriminant of the quadratic equation is negative ($\Delta < 0$), then the roots become imaginary due to square root of the negative discriminant. Hence, the two roots are unreal. Actually, the roots in algebraic form contain opposite signs. So, the roots of quadratic equation are complex conjugates.

The below example quadratic equation understands you how the roots of a quadratic equation are different imaginary numbers and complex conjugates.

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

Evaluate the discriminant of this quadratic equation.

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

$\implies \Delta = 49-120$

$\implies \Delta = -71$

The value of the discriminant of quadratic equation $5x^2+7x+6 = 0$ is a negative number. Now, find the roots of this equation.

$x \,=\, \dfrac{-7 \pm \sqrt{7^2-4 \times 5 \times 6}}{2 \times 5}$

$\implies$ $x \,=\, \dfrac{-7 \pm \sqrt{-71}}{10}$

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

Therefore, the roots $\dfrac{-7}{10}+i\dfrac{\sqrt{71}}{10}$ and $\dfrac{-7}{10}-i\dfrac{\sqrt{71}}{10}$ are different non-real numbers and complex conjugates.

It is proved that the roots of a quadratic equation are different imaginary and and complex conjugates if the discriminant of the quadratic equation is less than zero.

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