Question

How do solve \[{{x}^{3}}-4x-2=0\]?

Answer 1

Hint: This type of problem is based on the concept of solving cubic equations. First, we have to find the discriminant \[\Delta \] of the given equation. We find that \[\Delta >0\]. Therefore, the three roots are real and distinct. Then we need to use a trigonometric method to solve this type of question. Let us substitute \[x=k\cos \theta \] in the given equation. Expand the obtained equation and find the value of \[\theta \]. We then substitute \[\theta \] in x and make the necessary calculations to obtain the required solution of the given equation.

Complete step by step answer:
According to the question, we are asked to solve the given equation \[{{x}^{3}}-4x-2=0\].
We have given the equation is \[{{x}^{3}}-4x-2=0\] -----(1)
We first have to find the discriminant \[\Delta \].
\[\Delta ={{b}^{2}}{{c}^{2}}-4a{{c}^{3}}-4{{b}^{3}}d-27{{a}^{2}}{{d}^{2}}+18abcd\] for the cubic equation \[a{{x}^{2}}+bx+c=0\].
Here a=1, b=0, c=-4 and d=-2.
Substituting these values in \[\Delta \], we get
\[\Rightarrow \Delta =0{{\left( -4 \right)}^{2}}-4\left( 1 \right){{\left( -4 \right)}^{3}}-4\left( 0 \right)\left( -2 \right)-27\left( 1 \right){{\left( -2 \right)}^{2}}+12\left( 1 \right)\left( 0 \right)\left( -4 \right)\left( -2 \right)\]
On further simplification, we get,
\[\Rightarrow \Delta =0+4\left( 64 \right)+0-27\left( 4 \right)+0\]
\[\Rightarrow \Delta =0+256+0-108+0\]
\[\therefore \Delta =148\]
Here we observe that \[\Delta >0\]. Therefore, the roots are real and distinct.

Now substitute \[x=k\cos \theta \] in equation (1).
We get,
\[\Rightarrow {{\left( k\cos \theta \right)}^{3}}-4\left( k\cos \theta \right)-2=0\]
\[\Rightarrow {{k}^{3}}{{\cos }^{3}}\theta -4k\cos \theta -2=0\] --------(2)
Now substitute the value of k in (2) in such a way that we get \[4{{\cos }^{3}}\theta -3\cos \theta \] in equation (2).
Let \[k=\dfrac{4\sqrt{3}}{3}\]. Substituting in (2), we get
\[{{\left( \dfrac{4\sqrt{3}}{3} \right)}^{3}}{{\cos }^{3}}\theta -4\left( \dfrac{4\sqrt{3}}{3} \right)\cos \theta -2=0\]
\[\Rightarrow \left( \dfrac{{{4}^{3}}{{\left( \sqrt{3} \right)}^{3}}}{{{3}^{3}}} \right){{\cos }^{3}}\theta -\left( \dfrac{16\sqrt{3}}{3} \right)\cos \theta -2=0\]
On further simplification, we get,
\[\Rightarrow \left( \dfrac{64\left( 3 \right)\sqrt{3}}{27} \right){{\cos }^{3}}\theta -\left( \dfrac{16\sqrt{3}}{3} \right)\cos \theta -2=0\]
\[\Rightarrow \left( \dfrac{64\sqrt{3}}{9} \right){{\cos }^{3}}\theta -\left( \dfrac{16\sqrt{3}}{9} \right)3\cos \theta -2=0\] -------(3)
Take \[\dfrac{16\sqrt{3}}{9}\] common from equation (3). We get,
\[\left( \dfrac{16\sqrt{3}}{9} \right)\left( 4{{\cos }^{3}}\theta -3\cos \theta \right)-2=0\]
We know that \[\cos 3\theta =4{{\cos }^{3}}\theta -3\cos \theta \]. Therefore,
\[\left( \dfrac{16\sqrt{3}}{9} \right)\left( \cos 3\theta \right)-2=0\]
\[\Rightarrow \left( \dfrac{16\sqrt{3}}{9} \right)\left( \cos 3\theta \right)=2\]
\[\Rightarrow \cos 3\theta =2\left( \dfrac{9}{16\sqrt{3}} \right)\]
\[\Rightarrow \cos 3\theta =\dfrac{9}{8\sqrt{3}}\]
\[\therefore \cos 3\theta =\dfrac{3\sqrt{3}}{8}\] --------(4)

Now take \[{{\cos }^{-1}}\] on both sides of equation (4).
\[{{\cos }^{-1}}\left( \cos 3\theta \right)=\pm {{\cos }^{-1}}\left( \dfrac{3\sqrt{3}}{8} \right)+2k\pi \]
We know that \[{{\cos }^{-1}}\left( \cos \theta \right)=\theta \]. We get,
\[3\theta =\pm {{\cos }^{-1}}\left( \dfrac{3\sqrt{3}}{8} \right)+2k\pi \]
\[\Rightarrow \theta =\pm \dfrac{1}{3}{{\cos }^{-1}}\left( \dfrac{3\sqrt{3}}{8} \right)+\dfrac{2k\pi }{3}\]
Now take cos function on both sides of the obtained equation.
\[\Rightarrow \cos \theta =\cos \left( \dfrac{1}{3}{{\cos }^{-1}}\left( \dfrac{3\sqrt{3}}{8} \right)+\dfrac{2k\pi }{3} \right)\]
We know that \[x=k\cos \theta \] and \[k=\dfrac{4\sqrt{3}}{3}\]. Substituting these values in ‘x’, we get
\[x=\dfrac{4\sqrt{3}}{3}\cos \left( \dfrac{1}{3}{{\cos }^{-1}}\left( \dfrac{3\sqrt{3}}{8} \right)+\dfrac{2k\pi }{3} \right)\], where k=0,1,2.

Therefore, the three real and distinct values of ‘x’ are
\[x=\dfrac{4\sqrt{3}}{3}\cos \left( \dfrac{1}{3}{{\cos }^{-1}}\left( \dfrac{3\sqrt{3}}{8} \right) \right)\],
\[x=\dfrac{4\sqrt{3}}{3}\cos \left( \dfrac{1}{3}{{\cos }^{-1}}\left( \dfrac{3\sqrt{3}}{8} \right)+\dfrac{2\pi }{3} \right)\] and
\[\cos \theta x=\dfrac{4\sqrt{3}}{3}\cos \left( \dfrac{1}{3}{{\cos }^{-1}}\left( \dfrac{3\sqrt{3}}{8} \right)+\dfrac{4\pi }{3} \right)\]

Hence, the solutions of the given equation \[{{x}^{3}}-4x-2=0\] are\[\dfrac{4\sqrt{3}}{3}\cos \left( \dfrac{1}{3}{{\cos }^{-1}}\left( \dfrac{3\sqrt{3}}{8} \right) \right)\], \[\dfrac{4\sqrt{3}}{3}\cos \left( \dfrac{1}{3}{{\cos }^{-1}}\left( \dfrac{3\sqrt{3}}{8} \right)+\dfrac{2\pi }{3} \right)\] and\[\dfrac{4\sqrt{3}}{3}\cos \left( \dfrac{1}{3}{{\cos }^{-1}}\left( \dfrac{3\sqrt{3}}{8} \right)+\dfrac{4\pi }{3} \right)\] .

Note:
 Whenever you get this type of problem, we should always try to make the necessary calculations in the given equation to get the final of x which will be the required answer. We should avoid calculation mistakes based on sign conventions. We should be thorough with the trigonometric identities. It is always advisable to find the discriminant first and then solve the rest of the part.