Solving Equations of a Higher Degree

The number of roots or solutions [values for x] is dependent on the degree of the polynomial in an equation.  If the degree is 2, as in a quadratic equation, there is the possibility of two roots.  If the degree is six, there can be up to six roots. 

One method of solving equations of a higher degree is factoring.  The purpose is to factor the polynomial and reduce each factor to a linear or a quadratic factor.  If there is a quadratic factor than, use one of the four methods discussed in the post quadratic equations on this website.

Examples:         a)\: \: 4x^{4}-18x^{^{2}}=0

                                2x^{2}\left ( 2x^{2} -9\right )=0

                                2x^{2}=0\: \: or\: \: 2x^{2}=9

                                  x^{2}=0\: \: or\: \: x^{2}=\frac{9}{2}

                                  x=0\: \: or\: \: x=\pm \sqrt{\frac{9}{2}}

                                               x=\pm \frac{3}{\sqrt{2}}=\pm \frac{3\sqrt{2}}{2}

This example is of degree 4 and there were four roots.  One of the roots, 0, is a double root.  The other two roots are \left \{ \frac{3\sqrt{2}}{2},-\frac{3\sqrt{2}}{2} \right \}.

                  b)\: \: x^{3}-3x^{2}-x+3=0

                                   x^{2}\left ( x-3 \right )-1\left ( x-3 \right )=0

                                   \left ( x^{2}-1 \right )\left ( x-3 \right )=0

                                  \left ( x+1 \right )\left ( x-1 \right )\left ( x-3 \right )=0

                                   x=\left \{ -1,1,3 \right \}

                            c)\: \: 4x^{4}-65x^{2}+16=0

                                  \left ( 4x^{2}-1 \right )\left ( x^{2}-16 \right )=0

                                  4x^{2}=1\: \: \: \:\: \: \: \: \: \: \: x^{2}=16

                                     x^{2}=\frac{1}{4}\: \: \: \:\: \: \: \: \: \: \: x^{2}=16

                      x=\pm \frac{1}{2}\: \: \: \:\: \: \: \: \: \: \: x=\pm 4                          The 4 roots are \left \{ -\frac{1}{2} ,\frac{1}{2},-4,4\right \}.

Solving Radical Equations:

Isolate the radical and square or cube the expression to eliminate the radical.

Examples:        a)\: \: \sqrt{2x}-10=0

                                      \sqrt{2x}=10 

                                                             \left ( \sqrt{2x}\right )^{2}=\left ( 10 \right )^{2}

                                                                  2x=100

                                                                   x=50

                                          b)\: \: \sqrt{x-10}-4=0

                                                         \sqrt{x-10}=4

                                                             x-10=16

                                                                      x=26

                                            c)\: \: \sqrt[3]{2x+5}+3=0

                                                         \sqrt[3]{2x+5}=-3

                                     2x+5=-27       cube both sides of the equation   

                                            2x=-32    

                                                                            x=-16

                                             d)\: \: -\sqrt{26-11x}+4=x

                               \left ( 4-x_ \right )^{2}=\left ( \sqrt{26-11x} \right )^{2}          isolate the radical and square both sides

                             16-8x+x^{2}=26-11x

                             x^{2}+3x-10=0                       simplify the quadratic equation and factor

                            \left ( x+5 \right )\left ( x-2 \right )=0

                             x=\left \{ -5,2 \right \}

It is always wise to check your answers to see if they are valid.  In example d) both answers are correct.  When x=-5, the number under the root sign is +81 so -9+4= -5.  When x=2, the number under the root sign is +4 so -2+4=+2.  Both answers are valid.  If the number under the root sign was negative, then the answer would be an extraneous root.

                        e)\: \: \left ( x-5 \right )^{\frac{3}{2}}=8        

                            \sqrt[3]{\left ( x-5 \right )^{\frac{3}{2}}}=\sqrt[3]{2^{3}}      

                            \left ( x-5 \right )^{\frac{1}{2}}=2     

                             \sqrt{x-5}=2

                                x-5=4

                                     x=9                      Check:  \left ( 9-5 \right )^{\frac{3}{2}}=\sqrt{4^{3}}=\sqrt{64}=8

                   f)\: \: 3x\left ( x-1 \right )^{\frac{1}{2}}+2\left ( x-1 \right )^{\frac{3}{2}}=0

                      \left ( x-1 \right )^{\frac{1}{2}}\left [ 3x+2\left ( x-1 \right )^{\frac{2}{2}} \right ]=0          factor out the common factor \left ( x-1 \right )^{\frac{1}{2}}

                     \sqrt{x-1}=0\: \: or\: \: \left ( 3x+2x-2 \right )=0

                        x-1=0\: \: or\: \: 5x-2=0

                              x-1=0\: \: or\: \: 5x-2=0

                                    x=\left \{ 1,\frac{2}{5} \right \}            

               Check:  x=1,\: \: \: 0+0=0    valid

                             x=\frac{2}{5},   negative under the root sign, so not valid.

  Therefore the solution to example f) is only the one real number \rightarrow x=1.

Solving Rational Equations

To solve equations with fractions, the rule is to multiply by the LCD and thus eliminate the denominator.  Once the denominator has been eliminated solve using the strategies you have already learned for solving linear and quadratic equations.

Examples:        a)\: \: \frac{20-x}{x}=\frac{x}{1}                      

Since there is one fraction on each side of the equal sign, just cross multiply.

                              x^{2}=20-x

                             x^{2}+x-20=0

                             \left ( x+5 \right )\left ( x-4 \right )=0

                             x=\left \{ -5,4 \right \}              Check:   x=-5,\: \: \frac{20+5}{-5}=\frac{25}{-5}=-5

                                                                             x=4,\: \: \frac{20-4}{4}=\frac{16}{4}=4

As in radical equations, one must check the solutions to see if they are valid.  Any solution that makes the denominator equal to 0, is an invalid solution.

                   b)\: \: x=\frac{3}{x}+\frac{1}{2}

                        x=\frac{6+x}{2x}     The LCD is 2x 

                       2x^{2}=6+x                  

                      2x^{2}-x-6=0         

                     \left ( 2x+3 \right )\left ( x-2 \right )=0         

                         x=\left \{ -\frac{3}{2},2 \right \}      

                c)\: \: \frac{1}{x}-\frac{1}{x+1}=3         

                     x+1-x=3x\left ( x+1 \right )        the LCD is  x\left ( x+1 \right )

                             1=3x^{2}+3x     

                             3x^{2}+3x-1=0      using the quadratic formula  a=3, b=3, c= -1   

                            x=\frac{-3\pm \sqrt{9-4\left ( 3 \right )\left ( -1 \right )}}{6}=\frac{-3\pm \sqrt{9+12}}{6}=\frac{-3\pm \sqrt{21}}{6}          

                 d)\: \: \frac{4}{x+1}-\frac{3}{x+2}=1       the LCD is \left ( x+1 \right )\left ( x+2 \right )

                    4\left ( x+2 \right )-3\left ( x+1 \right )=\left ( x+1 \right )\left ( x+2 \right )

                    4x+8-3x-3=x^{2}+3x+2

                    x+5=x^{2}+3x+2

                    0=x^{2}+2x-3

                    0=\left ( x+3 \right )\left ( x-1 \right )

                   x=\left \{ -3,1 \right \}

Solving Absolute Value Equations

Absolute value equations have two solutions:

                                   \left | \square \right |=a,\: \: \rightarrow \square =a\: \: or\: \: \square =-a

Examples:           a)\: \: \left | 2x-1 \right |=5

                                   2x-1=-5          or         2x-1=5

                                        2x=-4                            2x=6

                                         x=-2                              x=3

                         b)\: \: \left | x \right |=x^{2}+x-3

                               x=-\left ( x^{2} +x-3\right )      or        x=\left ( x^{2} +x-3\right )

                              x=-x^{2}-x+3                      x=x^{2}+x-3

                              x^{2}+2x-3=0                        0=x^{2}-3

                             \left ( x+3 \right )\left ( x-1 \right )=0                    x^{2}=3

                              x=\left \{ -3,1 \right \}                             x=\left \{ \pm \sqrt{3} \right \}

                                                     x=\left \{ -3,-\sqrt{3},1,\sqrt{3} \right \}

                  c)\: \: \left | x+1 \right |=x^{2}-5

                       x+1=-\left ( x^{2}-5 \right )          or         x+1=\left ( x^{2}-5 \right )

                       x+1=-x^{2}+5                           x+1=x^{2}-5

                      x^{2}+x-4=0                               0=x^{2}-x-6

                      a = 1, b = 1, c = -4                        \left ( x-3 \right )\left ( x+2 \right )=0

                      x=\frac{-1\pm \sqrt{1-4\left ( 1 \right )\left ( -4 \right )}}{2}                  x=\left \{ 3,-2 \right \}

                      x=\frac{-1\pm \sqrt{1+16}}{2}=\frac{-1\pm \sqrt{17}}{2}

                                   x=\left \{\frac{-1+\sqrt{17}}{2}, \frac{-1-\sqrt{17}}{2},-2,3 \right \}

Filed under: Pre-requisites for Pre-Calculus

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