Question

# If the discriminant of the quadratic equation $a{{x}^{2}}+bx+c$ is zero, then its roots will be(a) real and equal (b) real and unequal (c) imaginary (d) zero

Hint: The discriminant of any quadratic equation is given by the formula:
$D={{b}^{2}}-4ac$
where b is the coefficient of x, a is the coefficient of ${{x}^{2}}$ and c is the constant term. The roots of the quadratic equation $a{{x}^{2}}+bx+c=0$ are given by the formula shown:
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
We will put the value of the discriminant is zero and then we will determine the nature of the roots obtained.

Before solving the question, we must know the terms given in the question like the quadratic equation and the discriminant of the quadratic equation. A quadratic equation is a type of equation in which the highest power of x present is 2. The general form of a quadratic equation is $a{{x}^{2}}+bx+c=0$. The discriminant of any quadratic equation is calculated by the formula:
$D={{b}^{2}}-4ac$
where b is the coefficient of x, a is the coefficient of ${{x}^{2}}$ and c is the constant term.
The roots of any quadratic equation are given by the formula shown:
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}....\left( i \right)$
Now, it is given in the question, that the discriminant is zero, i.e.
${{b}^{2}}-4ac=0....\left( ii \right)$
Now, we will put the value of ${{b}^{2}}-4ac$ from equation (ii) to equation (i). After doing this, we will get the following relation:
$x=\dfrac{-b\pm \sqrt{0}}{2a}$
Now, the value of $\sqrt{0}=0$, thus we get,
$x=\dfrac{-b\pm 0}{2a}$
$\Rightarrow {{x}_{1}}=\dfrac{-b+0}{2a}=\dfrac{-b}{2a}$
$\Rightarrow {{x}_{2}}=\dfrac{-b-0}{2a}=\dfrac{-b}{2a}$
Now, we can see that ${{x}_{1}}={{x}_{2}}$ but if the value of b and c both will be equal to zero, then the roots of the equation will be zero. Thus there are two possibilities either the roots are equal and non-zero or both the roots are zero.
Hence, options (a) and (d) are correct.

Note: We have assumed that all the numbers i.e. a, b, and c are real numbers. If the numbers a, b, and c will be imaginary then we cannot say that the roots will be real and equal or they will be zero. For example, if $a=\dfrac{i}{4},b=\sqrt{i}$ and c = i. Then, the roots will be
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
$x=\dfrac{-\sqrt{i}\pm \sqrt{{{\left( \sqrt{i} \right)}^{2}}-4\left( \dfrac{i}{4} \right)i}}{2\left( \dfrac{i}{4} \right)}$
$x=\dfrac{-\sqrt{i}\pm \sqrt{-1+1}}{\left( \dfrac{i}{2} \right)}$
$x=\dfrac{-\sqrt{i}\pm 0}{\left( \dfrac{i}{2} \right)}$
$\Rightarrow {{x}_{1}}=\dfrac{-2}{\sqrt{i}}$
$\Rightarrow {{x}_{2}}=\dfrac{-2}{\sqrt{i}}$
Thus, both the roots are the same, but they are imaginary.