Question

If the value of ${{b}^{2}}-4ac$ is equal to 0, the quadratic equation $a{{x}^{2}}+bx+c=0$ will have:
(a) Two real roots which are equal
(b) Two distinct real roots
(c) No real roots
(d) No roots or solutions

Answer 1

Hint: A quadratic equation $a{{x}^{2}}+bx+c=0$ is given and it is also given that ${{b}^{2}}-4ac$ is equal to 0 and are asked to find the information about the roots of this quadratic equation. We know that the formula for the roots of this quadratic equation is equal to: $x=\dfrac{-b\pm \sqrt{D}}{2a}$. In this formula, D is the discriminant and it’s value is equal to ${{b}^{2}}-4ac$ so we are going to put this as 0 in this formula for the roots of the equation and then see how the roots are varying.

Complete step by step solution:
We have given a quadratic equation as follows:
$a{{x}^{2}}+bx+c=0$
We have also given the relation of a, b and c as follows:
${{b}^{2}}-4ac=0$
We know that the formula for the roots of the quadratic equation $a{{x}^{2}}+bx+c=0$ is as follows:
$x=\dfrac{-b\pm \sqrt{D}}{2a}$
In this formula, D is the discriminant and the value of D is equal to ${{b}^{2}}-4ac$. Now, it is given that ${{b}^{2}}-4ac$ is 0 so the value of D is also 0 so substituting the value of D as 0 in the above value of x we get,
$x=\dfrac{-b\pm \sqrt{0}}{2a}$
$\Rightarrow x=\dfrac{-b\pm 0}{2a}$
Writing the above value of x when x is positive will give the value of x as:
$\begin{align}
  & \Rightarrow x=\dfrac{-b+0}{2a} \\
 & \Rightarrow x=\dfrac{-b}{2a} \\
\end{align}$
Writing the above value of x when we are taking the negative sign then the value of x is:
$\begin{align}
  & \Rightarrow x=\dfrac{-b-0}{2a} \\
 & \Rightarrow x=\dfrac{-b}{2a} \\
\end{align}$

So, the correct answer is “Option a”.

Note: From this problem, we have learnt an important concept in the quadratic equation that whenever the discriminant (D) value in the quadratic equation is 0 then the roots of this quadratic equation are real and equal. Using this concept we can find out the nature of roots.