Math Doubts

Nature of Roots of a Quadratic Equation if the Discriminant is positive

The roots of a quadratic equation are real numbers when the discriminant is positive.


The quadratic equation in standard algebraic form is $ax^2+bx+c = 0$ and its discriminant ($\Delta$ or $D$) is $b^2-4ac$.

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

If the quadratic equation has rational coefficients, and its discriminant is a positive ($\Delta > 0$) and cannot be expressed as a perfect square of a number, then the roots of the same quadratic equation are real and different irrational numbers.


$\sqrt{3}x^2+10x-8\sqrt{3} = 0$ is an example quadratic equation.

Calculate the discriminant of this quadratic equation.

$\Delta = 10^2-4 \times \sqrt{3} \times {(-8\sqrt{3})}$

$\implies \Delta = 100+96$

$\implies \Delta = 196$

The value of discriminant is a positive real number. It means $\Delta > 0$. Now, find the roots of this quadratic equation to study the nature of the roots.

$x \,=\, \dfrac{-10 \pm \sqrt{10^2-4 \times \sqrt{3} \times {(-8\sqrt{3})}}}{2 \times \sqrt{3}}$

$\implies$ $x \,=\, \dfrac{-10 \pm \sqrt{196}}{2\sqrt{3}}$

$\implies$ $x \,=\, \dfrac{-10 \pm 14}{2\sqrt{3}}$

$\implies$ $x \,=\, \dfrac{-10+14}{2\sqrt{3}}$ and $x \,=\, \dfrac{-10-14}{2\sqrt{3}}$

$\implies$ $x \,=\, \dfrac{4}{2\sqrt{3}}$ and $x \,=\, \dfrac{-24}{2\sqrt{3}}$

$\implies$ $\require{cancel} x \,=\, \dfrac{\cancel{4}}{\cancel{2}\sqrt{3}}$ and $\require{cancel} x \,=\, \dfrac{\cancel{-24}}{\cancel{2}\sqrt{3}}$

$\implies$ $x \,=\, \dfrac{2}{\sqrt{3}}$ and $x \,=\, \dfrac{-12}{\sqrt{3}}$

$\implies$ $x \,=\, \dfrac{2}{\sqrt{3}}$ and $x \,=\, \dfrac{-4 \times 3}{\sqrt{3}}$

$\implies$ $x \,=\, \dfrac{2}{\sqrt{3}}$ and $x \,=\, \dfrac{-4 \times {(\sqrt{3})}^2}{\sqrt{3}}$

$\implies$ $x \,=\, \dfrac{2}{\sqrt{3}}$ and $\require{cancel} x \,=\, \dfrac{-4 \times \cancel{{(\sqrt{3})}^2}}{\cancel{\sqrt{3}}}$

$\,\,\, \therefore \,\,\,\,\,\,$ $x \,=\, \dfrac{2}{\sqrt{3}}$ and $x \,=\, -4\sqrt{3}$

The roots of quadratic equation $\sqrt{3}x^2+10x-8\sqrt{3} = 0$ are $\dfrac{2}{\sqrt{3}}$ and $-4\sqrt{3}$. They are real numbers because the discriminant is positive and the value of square of discriminant is a real number. Therefore, it has proved that the roots of a quadratic equation are real if the value of discriminant is positive real number.

Follow us
Email subscription
Math Doubts
Math Doubts is a best place to learn mathematics and from basics to advanced scientific level for students, teachers and researchers. Know more
Follow us on Social Media
Mobile App for Android users Math Doubts Android App
Math Problems

Learn how to solve easy to difficult mathematics problems of all topics in various methods with step by step process and also maths questions for practising.

Learn more