Question

How do you find all the values of $x$ such that $f\left( x \right)=0$ given that $f\left( x \right)=\dfrac{{{x}^{3}}}{4}-2$?

Answer 1

Hint: Here we are given a cubic polynomial. It is a cubic polynomial as the highest degree of the given equation is 3. And we are asked to solve for $x$. Here again, we have to make use of basic mathematics as he has already provided us with the information that $f\left( x \right)=0$. All we have to do is just re-grouping some terms and find out the correct values of $x$ which satisfy this particular equation.

Complete step-by-step solution:
Solving for $x$ is nothing but finding out what values of $x$ satisfies the given equation. Indirectly we are asked to find out the roots of the equation. If we have a quadratic equation, the best we do to solve for $x$ or find out its root is either by splitting the middle term or by direction applying the formula which is $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.
But here, we have to make use of the equation $f\left( x \right)=0$.
So, let’s do that.
Upon equating $f\left( x \right)=0$, we get the following
$\begin{align}
  & \Rightarrow f\left( x \right)=0 \\
 & \Rightarrow f\left( x \right)=\dfrac{{{x}^{3}}}{4}-2=0 \\
 & \Rightarrow \dfrac{{{x}^{3}}}{4}-2=0 \\
\end{align}$
Let’s send $2$ to the right-hand side of the equation.
$\Rightarrow \dfrac{{{x}^{3}}}{4}=2$
Now let’s cross multiply the $4$. Upon doing so, we get the following
$\begin{align}
  & \Rightarrow \dfrac{{{x}^{3}}}{4}=2 \\
 & \Rightarrow {{x}^{3}}=4\times 2=8 \\
 & \Rightarrow {{x}^{3}}-8=0 \\
\end{align}$
So now we know that $x-2$ is a factor of $\dfrac{{{x}^{3}}}{4}-2$. It means that when $x-2$ divides $\dfrac{{{x}^{3}}}{4}-2$ , we get a remainder of 0.
We use the identity of ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)$.
${{x}^{3}}-8={{x}^{3}}-{{2}^{3}}=\left( x-2 \right)\left( {{x}^{2}}+2x+4 \right)$
But first, let’s look at the discriminant (D) for $\left( {{x}^{2}}+2x+4 \right)$.
We know $D=\sqrt{{{b}^{2}}-4ac}$ where $a$ is the coefficient of the ${{x}^{2}}$ term , $b$ is the coefficient of the $x$ term and $c$ is the constant term.
Upon comparing with our equations, we get
$\begin{align}
  & \Rightarrow a=1 \\
 & \Rightarrow b=2 \\
 & \Rightarrow c=4 \\
\end{align}$
Now, let’s plug it in the formula. Upon doing so, we get the following
$\begin{align}
  & \Rightarrow D=\sqrt{{{b}^{2}}-4ac}=\sqrt{{{2}^{2}}-4\times 1\times 4} \\
 & \Rightarrow D=\sqrt{4-16}<0 \\
\end{align}$
The discriminant is negative. It means there are no real values of $x$. $x$ takes up imaginary values here. The roots are $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.
So, for $\left( {{x}^{2}}+2x+4 \right)$, $x=\dfrac{-2\pm \sqrt{{{2}^{2}}-4\times 1\times 4}}{2\times 1}=\dfrac{-2\pm \sqrt{-12}}{2}=\dfrac{-2\pm 2i\sqrt{3}}{2}=-1\pm i\sqrt{3}$.
The value of $x$ in the function $f\left( x \right)=\dfrac{{{x}^{3}}}{4}-2$ takes up one real value which is 2. And two imaginary values.

Note: It is advisable to practice such questions. In doing so we can directly guess a value which will solve polynomials of degree more than 2. We should also be aware of the nature of discriminant. Whenever $D>0$, $D=0$ and $D<0$. Substituting the values of each in the formula carefully is also important as it may lead to calculation errors and in the end, we may get wrong answers.