Question

How do you solve $3{{x}^{4}}+12{{x}^{2}}-15=0$?

Answer 1

Hint: Firstly, the given equation of fourth degree has to be converted to the quadratic equation by substituting $y={{x}^{2}}$ into the equation so that the given equation will become $3{{y}^{2}}+12y-5=0$. This equation can be solved by using the quadratic formula which is given by $y=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. In the quadratic equation $3{{y}^{2}}+12y-5=0$ we can note $a=3,b=12,c=-15$. Therefore, on substituting these values into the quadratic formula, we will obtain the two values for y. Then, putting $y={{x}^{2}}$ we will obtain the four values for x and hence the solutions for the given equation.

Complete step-by-step answer:
The equation given in the above question is
$\begin{align}
  & \Rightarrow 3{{x}^{4}}+12{{x}^{2}}-15=0 \\
 & \Rightarrow 3{{\left( {{x}^{2}} \right)}^{2}}+12{{x}^{2}}-15=0 \\
\end{align}$
Let us put $y={{x}^{2}}$ into the above equation to get
$\Rightarrow 3{{y}^{2}}+12y-15=0$
From the above equation, we can note the values of the coefficients as
$\begin{align}
  & \Rightarrow a=3 \\
 & \Rightarrow b=12 \\
 & \Rightarrow c=-15 \\
\end{align}$
We know that the solution of a quadratic equation is given by the quadratic formula, which is given as
$\Rightarrow y=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Substituting the values of the coefficients into the above formula, we get
\[\begin{align}
  & \Rightarrow y=\dfrac{-\left( 12 \right)\pm \sqrt{{{\left( 12 \right)}^{2}}-4\left( 3 \right)\left( -15 \right)}}{2\left( 3 \right)} \\
 & \Rightarrow y=\dfrac{-12\pm \sqrt{144+180}}{6} \\
 & \Rightarrow y=\dfrac{-12\pm \sqrt{324}}{6} \\
\end{align}\]
We know that \[\sqrt{324}=18\]. Putting it above, we get
\[\begin{align}
  & \Rightarrow y=\dfrac{-12\pm 18}{6} \\
 & \Rightarrow y=\dfrac{-12+18}{6},y=\dfrac{-12-18}{6} \\
 & \Rightarrow y=\dfrac{6}{6},y=\dfrac{-30}{6} \\
 & \Rightarrow y=1,y=-5 \\
\end{align}\]
Now, according to our substitution, $y={{x}^{2}}$. Therefore, we get
$\begin{align}
  & \Rightarrow {{x}^{2}}=1,{{x}^{2}}=-5 \\
 & \Rightarrow x=\pm 1,x=\pm \sqrt{5}i \\
\end{align}$
Hence, the solutions of the given equation are $x=1,x=-1,x=\sqrt{5}i,x=-\sqrt{5}i$.

Note: Although the given equation is a fourth degree equation, it can be converted in the form of a quadratic equation, as done in the above solution. We must always try for this whenever we have a higher degree equation. Do not end the solution at the obtained values of y. This is because the variable in the original equation was x, and not y.