The nature of roots of a quadratic equation are real and distinct irrational numbers when the quadratic equation has rational coefficients and its discriminant is positive and not a perfect square number.
$a$, $b$ and $c$ are rational coefficients of a quadratic equation $ax^2+bx+c = 0$ and the discriminant ($\Delta$) of this equation 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.
The below quadratic equation is an example quadratic equation which contains rational numbers and coefficients. Moreover, the discriminant is a positive number and cannot be written as a perfect square number.
$2x^2+5x-6 = 0$
$2$, $5$ and $-6$ are rational numbers and they are coefficients of this quadratic equation. Now, find the discriminant of the quadratic equation.
$\Delta = 5^2-4 \times 2 \times (-6)$
$\implies \Delta = 25+48$
$\implies \Delta = 73$
The discriminant cannot be written as a square of a number. Calculate the roots of this example quadratic equation.
$x \,=\, \dfrac{-5 \pm \sqrt{5^2-4 \times 2 \times (-6)}}{2 \times 2}$
$\implies$ $x \,=\, \dfrac{-5 \pm \sqrt{73}}{4}$
$\,\,\, \therefore \,\,\,\,\,\,$ $x \,=\, \dfrac{-5+\sqrt{73}}{4}$ and $x \,=\, \dfrac{-5-\sqrt{73}}{4}$
The roots of quadratic equation $2x^2+5x-6 = 0$ are $\dfrac{-5+\sqrt{73}}{4}$ and $\dfrac{-5-\sqrt{73}}{4}$. The two roots are real but irrational because the discriminant is not a perfect square of a number.
Therefore, it has proved that the roots of a quadratic equation having rational coefficients are different irrational numbers when the value of discriminant is positive and cannot be expressed as a square number.
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