How to solve a Cubic Equation? 3 Solved Examples

The basic approach for solving a cubic equation is to shrink it to a quadratic equation and then try to solve the quadratic equation by adopting the general procedure like by factorizing or by the quadratic formula.

Generally, all cubic equations have either one or three real roots. A general form of cubic equation is given by,

\displaystyle a{{x}^{3}}+b{{x}^{2}}+cx+d=0

Some examples of cubic equations are;

\displaystyle 2{{x}^{3}}+5{{x}^{2}}+2x+4=0

\displaystyle 3{{x}^{3}}+2{{x}^{2}}-6x+24=0

\displaystyle 8{{x}^{3}}+27=0

\displaystyle {{x}^{3}}-64=0

In order to solve the cubic equations, we recall the quadratic equation which can easily be solved by the quadratic formula,

\displaystyle \frac{{-b\pm \sqrt{{{{b}^{2}}-4ac}}}}{{2a}}

Where, \displaystyle {{b}^{2}}-4ac is discriminant of the quadratic which is denoted by a Greek letter Delta (\displaystyle \Delta ). There are possible three cases;

  • If \displaystyle \Delta =0 the quadratic equation has one real solution.
  • If \displaystyle \Delta >0 then the quadratic equation has two real solutions.
  • If \displaystyle \Delta <0 then the quadratic equation has no real solutions.

Now the question rises here that how to find the real solution of a cubic equation, so, let’s start

First we consider,

\displaystyle p=b-\frac{{{{a}^{2}}}}{3}\,\,\,\,and\,\,\,q=\frac{{2{{a}^{3}}}}{{27}}-\frac{{ab}}{3}+c

And then we will find the discriminant i.e. \displaystyle \Delta as given by

\displaystyle \Delta =\frac{{{{q}^{2}}}}{4}+\frac{{{{p}^{3}}}}{{27}}

If \displaystyle \Delta =0 then there are repeated roots as follows

\displaystyle {{x}_{1}}=-2{{\left( {\frac{q}{2}} \right)}^{{\frac{1}{3}}}}-\frac{a}{3}\,\,\,and\,\,\,{{x}_{2}}={{x}_{3}}={{\left( {\frac{q}{2}} \right)}^{{\frac{1}{3}}}}-\frac{a}{3}

If \displaystyle \Delta >0 then there is one real solution as follows,

\displaystyle x={{\left( {-\frac{q}{2}+\sqrt{\Delta }} \right)}^{{\frac{1}{3}}}}+{{\left( {-\frac{q}{2}-\sqrt{\Delta }} \right)}^{{\frac{1}{3}}}}-\frac{a}{3}

If \displaystyle \Delta <0 then there are three real solution as follows,

\displaystyle {{x}_{1}}=\frac{2}{{\sqrt{3}}}\sqrt{{-p}}\sin \left( {\frac{1}{3}{{{\sin }}^{{-1}}}\left( {\frac{{3\sqrt{3}q}}{{2{{{(\sqrt{{-p}})}}^{3}}}}} \right)} \right)-\frac{a}{3}\,\,

\displaystyle {{x}_{2}}=\frac{2}{{\sqrt{3}}}\sqrt{{-p}}\sin \left( {\frac{1}{3}{{{\sin }}^{{-1}}}\left( {\frac{{3\sqrt{3}q}}{{2{{{(\sqrt{{-p}})}}^{3}}}}+\frac{\pi }{3}} \right)} \right)-\frac{a}{3}\,\,

\displaystyle {{x}_{3}}=\frac{2}{{\sqrt{3}}}\sqrt{{-p}}\sin \left( {\frac{1}{3}{{{\sin }}^{{-1}}}\left( {\frac{{3\sqrt{3}q}}{{2{{{(\sqrt{{-p}})}}^{3}}}}+\frac{\pi }{6}} \right)} \right)-\frac{a}{3}

Example No.1       

Find the real solution of following cubic equation?

\displaystyle {{x}^{3}}+4{{x}^{2}}-2x+12=0

Solution:

\displaystyle {{x}^{3}}+4{{x}^{2}}-2x+12=0

Here, \displaystyle a=4\,,\,\,b=-2\,,\,\,c=12

\displaystyle p=b-\frac{{{{a}^{2}}}}{3}\,\,=-2-\frac{{16}}{3}=\frac{{-6-16}}{3}\,\,

\displaystyle =\frac{{-22}}{3}=-7.33

\displaystyle q=\frac{{2{{a}^{3}}}}{{27}}-\frac{{ab}}{3}+c=\frac{{2\times {{4}^{3}}}}{{27}}-\frac{{4\times (-2)}}{3}+12

\displaystyle =\frac{{128}}{{27}}+\frac{8}{3}+12=\frac{{128+72+324}}{{27}}

\displaystyle =\frac{{624}}{{27}}=23.11

\displaystyle \Delta =\frac{{{{q}^{2}}}}{4}+\frac{{{{p}^{3}}}}{{27}}

\displaystyle \Delta =\frac{{{{{(23.11)}}^{2}}}}{4}+\frac{{{{{(-7.33)}}^{3}}}}{{27}}

\displaystyle =133.51-14.586=118.924

Here, \displaystyle \Delta is greater than 0, therefore, there is only one real solution which is given by,

\displaystyle x={{\left( {-\frac{q}{2}+\sqrt{\Delta }} \right)}^{{\frac{1}{3}}}}+{{\left( {-\frac{q}{2}-\sqrt{\Delta }} \right)}^{{\frac{1}{3}}}}-\frac{a}{3}

\displaystyle x={{\left( {-\frac{{23.11}}{2}+\sqrt{{118.924}}} \right)}^{{\frac{1}{3}}}}+{{\left( {-\frac{{23.11}}{2}-\sqrt{{118.924}}} \right)}^{{\frac{1}{3}}}}-\frac{4}{3}

\displaystyle x={{\left( {-11.555+10.905} \right)}^{{\frac{1}{3}}}}+{{\left( {-11.555-10.905} \right)}^{{\frac{1}{3}}}}-\frac{4}{3}

\displaystyle x={{\left( {-0.65} \right)}^{{\frac{1}{3}}}}+{{\left( {-22.46} \right)}^{{\frac{1}{3}}}}-\frac{4}{3}

\displaystyle x=-0.866-2.82-1.333

\displaystyle x=-5.019

Example No.2

Find the real solution of following cubic equation?

\displaystyle {{x}^{3}}+{{x}^{2}}+x+4=0

Solution:

\displaystyle {{x}^{3}}+{{x}^{2}}+x+4=0

Where,

\displaystyle a=1\,,\,\,b=1\,,\,\,c=4

\displaystyle p=b-\frac{{{{a}^{2}}}}{3}\,\,=1-\frac{1}{3}=\frac{{3-1}}{3}\,\,

\displaystyle =\frac{2}{3}=0.667

\displaystyle q=\frac{{2{{a}^{3}}}}{{27}}-\frac{{ab}}{3}+c=\frac{{2\times {{1}^{3}}}}{{27}}-\frac{{1\times 1}}{3}+4

\displaystyle =\frac{2}{{27}}-\frac{2}{3}+4=\frac{{2-18+108}}{{27}}

\displaystyle =\frac{{92}}{{27}}=3.407

\displaystyle \Delta =\frac{{{{q}^{2}}}}{4}+\frac{{{{p}^{3}}}}{{27}}

\displaystyle \Delta =\frac{{{{{(3.407)}}^{2}}}}{4}+\frac{{{{{(0.667)}}^{3}}}}{{27}}

\displaystyle =2.90+0.011=2.911

Here, \displaystyle \Delta is again greater than 0, therefore, there is only one real solution which is given by,

\displaystyle x={{\left( {-\frac{q}{2}+\sqrt{\Delta }} \right)}^{{\frac{1}{3}}}}+{{\left( {-\frac{q}{2}-\sqrt{\Delta }} \right)}^{{\frac{1}{3}}}}-\frac{a}{3}

\displaystyle x={{\left( {-\frac{{3.407}}{2}+\sqrt{{2.911}}} \right)}^{{\frac{1}{3}}}}+{{\left( {-\frac{{3.407}}{2}-\sqrt{{2.911}}} \right)}^{{\frac{1}{3}}}}-\frac{1}{3}

\displaystyle x={{\left( {-1.703+1.71} \right)}^{{\frac{1}{3}}}}+{{\left( {-1.703-1.71} \right)}^{{\frac{1}{3}}}}-\frac{1}{3}

\displaystyle x={{\left( {0.007} \right)}^{{\frac{1}{3}}}}+{{\left( {9.413} \right)}^{{\frac{1}{3}}}}-\frac{1}{3}

\displaystyle x=0.191-2.111-0.333

\displaystyle x=-2.253

Example No.3

Find the real solution of following cubic equation?

\displaystyle {{x}^{3}}+3{{x}^{2}}+4x+5=0

Solution:

\displaystyle {{x}^{3}}+3{{x}^{2}}+4x+5=0

\displaystyle a=3\,,\,\,b=4\,,\,\,c=5

\displaystyle p=b-\frac{{{{a}^{2}}}}{3}\,\,=4-\frac{9}{3}=\frac{{12-9}}{3}\,\,

\displaystyle =\frac{3}{3}=1

\displaystyle q=\frac{{2{{a}^{3}}}}{{27}}-\frac{{ab}}{3}+c=\frac{{2\times {{3}^{3}}}}{{27}}-\frac{{3\times 4}}{3}+5

\displaystyle =\frac{{54}}{{27}}-\frac{{12}}{3}+5=\frac{{54-108+135}}{{27}}

\displaystyle =\frac{{81}}{{27}}=3

\displaystyle \Delta =\frac{{{{q}^{2}}}}{4}+\frac{{{{p}^{3}}}}{{27}}

\displaystyle \Delta =\frac{{{{{(3)}}^{2}}}}{4}+\frac{{{{{(1)}}^{3}}}}{{27}}

\displaystyle =2.25+0.037=2.287

Again here, \displaystyle \Delta is again greater than 0, therefore, there is only one real solution which is given by,

\displaystyle x={{\left( {-\frac{q}{2}+\sqrt{\Delta }} \right)}^{{\frac{1}{3}}}}+{{\left( {-\frac{q}{2}-\sqrt{\Delta }} \right)}^{{\frac{1}{3}}}}-\frac{a}{3}

\displaystyle x={{\left( {-\frac{3}{2}+\sqrt{{2.287}}} \right)}^{{\frac{1}{3}}}}+{{\left( {-\frac{3}{2}-\sqrt{{2.287}}} \right)}^{{\frac{1}{3}}}}-\frac{3}{3}

\displaystyle x={{\left( {-1.5+1.512} \right)}^{{\frac{1}{3}}}}+{{\left( {-1.5-1.512} \right)}^{{\frac{1}{3}}}}-1

\displaystyle x={{\left( {0.012} \right)}^{{\frac{1}{3}}}}+{{\left( {-3.012} \right)}^{{\frac{1}{3}}}}-1

\displaystyle x=0.229-1.444-1

\displaystyle x=-2.215