Question

Consider the cubic equation ${{x}^{3}}+p{{x}^{2}}+qx+r=0$ , where $p$ , $q$ , $r$ are the real numbers. Which of the following statements is correct?
A. If ${{p}^{2}}-2q<0$ , then the equation has one real and two imaginary roots.
B. If ${{p}^{2}}-2q\ge 0$, then the equation has all real roots.
 C. If ${{p}^{2}}-2q>0$ , then the equation has all real and distinct roots.
D. If $4{{p}^{3}}-27{{q}^{2}}>0$, then the equation has real and distinct roots.

Answer 1

Hint: In this problem we need to check for the correct statement from the given set of statements. We will first consider the given cubic equation as $f\left( x \right)={{x}^{3}}+p{{x}^{2}}+qx+r$ . Now we will differentiate the equation with respect to $x$ and calculate the value of ${{f}^{'}}\left( x \right)$ . For the obtained equation we will calculate the value of discriminant and analyze the discriminant value by considering all the given statements.

Complete step by step answer:
The cubic equation is ${{x}^{3}}+p{{x}^{2}}+qx+r=0$. Where $p$ , $q$ , $r$ are the real numbers.
Considering the given cubic equation as $f\left( x \right)={{x}^{3}}+p{{x}^{2}}+qx+r$.
Differentiating the above equation with respect to $x$ , then we will have
${{f}^{'}}\left( x \right)=\dfrac{d}{dx}\left( {{x}^{3}}+p{{x}^{2}}+qx+r \right)$
Applying differentiation for each term individually, then we will get
${{f}^{'}}\left( x \right)=\dfrac{d}{dx}\left( {{x}^{3}} \right)+p\dfrac{d}{dx}\left( {{x}^{2}} \right)+q\dfrac{d}{dx}\left( x \right)+\dfrac{d}{dx}\left( r \right)$
Applying the differentiation formulas $\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}$ , $\dfrac{d}{dx}\left( \text{constant} \right)=0$ in the above equation, then we will have
$\begin{align}
  & {{f}^{'}}\left( x \right)=3{{x}^{3-1}}+p\left( 2{{x}^{2-1}} \right)+q\left( 1{{x}^{1-1}} \right)+0 \\
 & \Rightarrow {{f}^{'}}\left( x \right)=3{{x}^{2}}+2px+q \\
\end{align}$
We have the derivative of the given cubic equation as $3{{x}^{2}}+2px+q$ which is a quadratic equation.
The discriminant of the quadratic equation is given by
$\begin{align}
  & {{b}^{2}}-4ac={{\left( 2p \right)}^{2}}-4\left( 3 \right)\left( q \right) \\
 & \Rightarrow {{b}^{2}}-4ac=4{{p}^{2}}-12q \\
\end{align}$
Taking $4$ as common from the above equation, then we will have
$\begin{align}
  & {{b}^{2}}-4ac=4\left( {{p}^{2}}-3q \right) \\
 & \Rightarrow {{b}^{2}}-4ac=4\left( {{p}^{2}}-2q-q \right) \\
\end{align}$
From the above value we can say that If ${{p}^{2}}-2q<0$ , then the equation has one real and two imaginary roots.
Since the above discriminant will become a negative value results in two imaginary roots for the given cubic equation.

So, the correct answer is “Option A”.

Note: In this problem we have given the cubic equation and asked to find the nature of the roots. In some cases they may give a quadratic equation and ask to find the nature of the roots. Then also we will calculate the value of discriminant and based on the value of discriminant we will decide the nature of the roots.
If discriminant is a positive value, then the equation has real and distinct roots.
If the discriminant is a negative value, then the equation has complex and distinct roots.
If the discriminant is equal to zero, then the equation has real and equal roots.