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.

Latest Math Topics

Jul 20, 2023

Jun 26, 2023

Jun 23, 2023

Latest Math Problems

Oct 15, 2023

Copyright © 2012 - 2023 Math Doubts, All Rights Reserved