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.

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