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

Answer 1

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.

Complete step-by-step answer:
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.