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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