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Question: If the discriminant of the quadratic equation \[a{{x}^{2}}+bx+c\] is zero, then its roots will be ...

If the discriminant of the quadratic equation ax2+bx+ca{{x}^{2}}+bx+c is zero, then its roots will be
(a) real and equal
(b) real and unequal
(c) imaginary
(d) zero

Explanation

Solution

Hint: The discriminant of any quadratic equation is given by the formula:
D=b2−4acD={{b}^{2}}-4ac
where b is the coefficient of x, a is the coefficient of x2{{x}^{2}} and c is the constant term. The roots of the quadratic equation ax2+bx+c=0a{{x}^{2}}+bx+c=0 are given by the formula shown:
x=−b±b2−4ac2ax=\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 ax2+bx+c=0a{{x}^{2}}+bx+c=0. The discriminant of any quadratic equation is calculated by the formula:
D=b2−4acD={{b}^{2}}-4ac
where b is the coefficient of x, a is the coefficient of x2{{x}^{2}} and c is the constant term.
The roots of any quadratic equation are given by the formula shown:
x=−b±b2−4ac2a....(i)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.
b2−4ac=0....(ii){{b}^{2}}-4ac=0....\left( ii \right)
Now, we will put the value of b2−4ac{{b}^{2}}-4ac from equation (ii) to equation (i). After doing this, we will get the following relation:
x=−b±02ax=\dfrac{-b\pm \sqrt{0}}{2a}
Now, the value of 0=0\sqrt{0}=0, thus we get,
x=−b±02ax=\dfrac{-b\pm 0}{2a}
⇒x1=−b+02a=−b2a\Rightarrow {{x}_{1}}=\dfrac{-b+0}{2a}=\dfrac{-b}{2a}
⇒x2=−b−02a=−b2a\Rightarrow {{x}_{2}}=\dfrac{-b-0}{2a}=\dfrac{-b}{2a}
Now, we can see that x1=x2{{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=i4,b=ia=\dfrac{i}{4},b=\sqrt{i} and c = i. Then, the roots will be
x=−b±b2−4ac2ax=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}
x=−i±(i)2−4(i4)i2(i4)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=−i±−1+1(i2)x=\dfrac{-\sqrt{i}\pm \sqrt{-1+1}}{\left( \dfrac{i}{2} \right)}
x=−i±0(i2)x=\dfrac{-\sqrt{i}\pm 0}{\left( \dfrac{i}{2} \right)}
⇒x1=−2i\Rightarrow {{x}_{1}}=\dfrac{-2}{\sqrt{i}}
⇒x2=−2i\Rightarrow {{x}_{2}}=\dfrac{-2}{\sqrt{i}}
Thus, both the roots are the same, but they are imaginary.