The roots of a quadratic equation are real and also a repeated or double root. It is only possible when the discriminant of a quadratic equation is equal to zero.

$ax^2+bx+c = 0$ is a quadratic equation and its discriminant ($\Delta$) is $b^2-4ac$.

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

If the discriminant of the quadratic equation is zero ($\Delta = 0$), then the roots are $\dfrac{-b + \sqrt{0}}{2a}$ and $\dfrac{-b \,-\sqrt{0}}{2a}$.

Therefore, the two roots are $\dfrac{-b}{2a}$ and $\dfrac{-b}{2a}$. In this case, the two roots are same and equal.

Actually, the literals $a$, $b$ and $c$ are real numbers. So, the two roots should also be real numbers.

The property of the roots can be understood from an example quadratic equation.

$9x^2+30x+25 = 0$

Find the discriminant of the quadratic equation.

$\Delta = 30^2-4 \times 9 \times 25$

$\implies \Delta = 900-900$

$\implies \Delta = 0$

The value of the discriminant of the quadratic equation $9x^2+30x+25 = 0$ is zero. Now, evaluate the roots of this equation.

$x \,=\, \dfrac{-30 \pm \sqrt{30^2-4 \times 9 \times 25}}{2 \times 9}$

$\implies$ $x \,=\, \dfrac{-30 \pm \sqrt{0}}{18}$

$\implies$ $x \,=\, \dfrac{-30+0}{18}$ and $x \,=\, \dfrac{-30-0}{18}$

$\implies$ $x \,=\, \dfrac{-30}{18}$ and $x \,=\, \dfrac{-30}{18}$

$\,\,\, \therefore \,\,\,\,\,\,$ $x \,=\, -\dfrac{5}{3}$ and $x \,=\, -\dfrac{5}{3}$

Therefore $x \,=\, -\dfrac{5}{3}$ and $-\dfrac{5}{3}$ are the two roots of this quadratic equation. The two roots are equal and real numbers. Hence, they are called repeated or double root simply.

It is proved that the roots of a quadratic equation are real and equal when the discriminant of the quadratic equation is equal to zero.

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