The roots of a quadratic equation are real and equal if the discriminant of a quadratic equation is zero.

In mathematics, the quadratic equation in standard form is written as $ax^2+bx+c \,=\, 0$ algebraically. The discriminant ($\Delta$ or $D$) of this quadratic equation is written as $b^2-4ac$.

The roots (or zeros) of the quadratic equation in terms of discriminant are written in the following two forms.

$(1).\,\,\,$ $\dfrac{-b+\sqrt{\Delta}}{2a}$

$(2).\,\,\,$ $\dfrac{-b-\sqrt{\Delta}}{2a}$

If the discriminant of the quadratic equation is zero, then the square root of the discriminant will also be positive.

$\implies$ $\sqrt{\Delta} \,=\, 0$

The zeros or roots of the quadratic equation can be expressed in the following two mathematical forms.

$(1).\,\,\,$ $\dfrac{-b+0}{2a}$ $\,=\,$ $\dfrac{-b}{2a}$

$(2).\,\,\,$ $\dfrac{-b-0}{2a}$ $\,=\,$ $\dfrac{-b}{2a}$

The two roots express clearly that the roots of the quadratic equation are real and same.

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

Let’s evaluate the discriminant for the given quadratic equation.

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

$\implies$ $\Delta \,=\, 900-900$

$\implies$ $\Delta \,=\, 0$

Now, we can evaluate the square root of the discriminant.

$\implies$ $\sqrt{\Delta} \,=\, \sqrt{0}$

$\implies$ $\sqrt{\Delta} \,=\, 0$

We can now evaluate the roots for the given quadratic equation.

$(1).\,\,\,$ $\dfrac{-30+0}{2 \times 9}$ $\,=\,$ $\dfrac{-30}{18}$ $\,=\,$ $-\dfrac{5}{3}$

$(2).\,\,\,$ $\dfrac{-30-0}{2 \times 9}$ $\,=\,$ $\dfrac{-30}{18}$ $\,=\,$ $-\dfrac{5}{3}$

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

Therefore, it is proved that the roots are equal and also real numbers when the discriminant of quadratic equation is zero.

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