The roots of a quadratic equation are real and distinct if the discriminant of a quadratic equation is positive.

Algebraically, a quadratic equation is written as $ax^2+bx+c \,=\, 0$ in mathematical form. Now, the discriminant ($\Delta$ or $D$) of this quadratic equation is written as $b^2-4ac$.

The zeros or roots of the quadratic equation in terms of discriminant are written in the following two forms.

$(1).\,\,\,$ $\dfrac{-b+\sqrt{\Delta}}{2a}$

$(2).\,\,\,$ $\dfrac{-b-\sqrt{\Delta}}{2a}$

When the discriminant of the quadratic equation is positive, the square root of the discriminant will be positive too.

For example $\sqrt{\Delta} \,=\, d$

The zeros or roots of the quadratic equation can be expressed in the following two mathematical forms.

$(1).\,\,\,$ $\dfrac{-b+d}{2a}$

$(2).\,\,\,$ $\dfrac{-b-d}{2a}$

The two roots clearly express that the roots of the quadratic equation are real and distinct.

$2x^2+13x+15 = 0$

Let’s evaluate the discriminant of the given quadratic equation.

$\Delta \,=\, 13^2-4 \times 2 \times 15$

$\implies$ $\Delta \,=\, 169-120$

$\implies$ $\Delta \,=\, 49$

Now, find the square root of the discriminant.

$\implies$ $\sqrt{\Delta} \,=\, \sqrt{49}$

$\implies$ $\sqrt{\Delta} \,=\, 7$

We can now evaluate the roots for the given quadratic equation.

$(1).\,\,\,$ $\dfrac{-13+7}{2 \times 2}$ $\,=\,$ $\dfrac{-6}{4}$ $\,=\,$ $-\dfrac{3}{2}$

$(2).\,\,\,$ $\dfrac{-13-7}{2 \times 2}$ $\,=\,$ $\dfrac{-20}{4}$ $\,=\,$ $-5$

$\,\,\, \therefore \,\,\,\,\,\,$ $x \,=\, -\dfrac{3}{2}$ and $x \,=\, -5$

It is proved that the roots are distinct and also real numbers when the discriminant of quadratic equation is greater than zero.

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