Nature of Roots of a Quadratic Equation for Square discriminant

The roots of a quadratic equation are real and rational if the discriminant of a quadratic equation is positive and a perfect square.

Introduction

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.

Example

The following quadratic equation is an example for a quadratic equation which contains rational numbers as coefficients and the discriminant is positive number, can expressed as a perfect square number.

$3x^2-10x+3 = 0$

$3$, $-10$ and $3$ are rational numbers and they are coefficients of this quadratic equation. Now, find the discriminant of the quadratic equation.

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

$\implies \Delta = 100-36$

$\implies \Delta = 64$

$\implies \Delta = 8^2$

The discriminant is expressed as a square of a number. Now, evaluate the roots of this quadratic equation.

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

$\implies$ $x \,=\, \dfrac{10 \pm \sqrt{64}}{6}$

$\implies$ $x \,=\, \dfrac{10 \pm \sqrt{8^2}}{6}$

$\implies$ $x \,=\, \dfrac{10 \pm 8}{6}$

$\implies$ $x \,=\, \dfrac{10+8}{6}$ and $\dfrac{10-8}{6}$

$\implies$ $x \,=\, \dfrac{18}{6}$ and $\dfrac{2}{6}$

$\implies$ $\require{cancel} x \,=\, \dfrac{\cancel{18}}{\cancel{6}}$ and $\dfrac{\cancel{2}}{\cancel{6}}$

$\,\,\, \therefore \,\,\,\,\,\,$ $x \,=\, 3$ and $x \,=\, \dfrac{1}{3}$

The roots of the quadratic equation $3x^2-10x+3 = 0$ are $3$ and $\dfrac{1}{3}$, they are real numbers but rational because the discriminant is a perfect square of a number.

Therefore, it is proved that the roots of a quadratic equation are distinct rational numbers when the discriminant is a positive number, which can be expressed as a square of another number.

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