Functions: Power functions
Equations with power functions
In quadratic equations we have seen how to solve an equation #x^2=c#. With the same procedure, we will use higher degree roots to solve an equation #x^n=c#.
The solutions to the equation #x^\orange{n}=\blue{c}# are dependent on the values of #\orange n# and #\blue c#.
#\blue{c} \gt 0# | #\blue{c}=0# | #\blue{c} \lt 0# | |
#\orange{n}# is even |
Two solutions: #x=-\sqrt[\orange{n}]{\blue{c}} \lor x=\sqrt[\orange{n}]{\blue{c}}# |
One solution: #x=0# |
No solutions
|
#\orange{n}# is odd |
One solution: #x=\sqrt[\orange{n}]{\blue{c}}# |
One solution: #x=0# |
One solution: #x=\sqrt[\orange{n}]{\blue{c}}# |
In the examples we see that you can reduce many equations to the form #x^\orange{n}=\blue{c}# and then solve them.
#x=\frac{1}{6}\sqrt[4]{1512} \lor x=-\frac{1}{6}\sqrt[4]{1512}#
#\begin{array}{rcl}6\, x^{4}+2&=& 9 \\
&&\phantom{xxx}\blue{\text{the equation we need to solve}} \\
6\, x^{4}&=&7 \\
&&\phantom{xxx}\blue{\text{both sides minus }2} \\
x^{4} &=& {{7}\over{6}} \\
&&\phantom{xxx}\blue{\text{both sides divided by }6} \\
x=\sqrt[4]{{{7}\over{6}}} &\lor& x=-\sqrt[4]{{{7}\over{6}}} \\
&&\phantom{xxx}\blue{\text{both sides taken the }4 \text{-th root}}\\
x=\frac{1}{6}\sqrt[4]{1512} &\lor& x=-\frac{1}{6}\sqrt[4]{1512}\\
&&\phantom{xxx}\blue{\text{simplified}} \end{array}#
#\begin{array}{rcl}6\, x^{4}+2&=& 9 \\
&&\phantom{xxx}\blue{\text{the equation we need to solve}} \\
6\, x^{4}&=&7 \\
&&\phantom{xxx}\blue{\text{both sides minus }2} \\
x^{4} &=& {{7}\over{6}} \\
&&\phantom{xxx}\blue{\text{both sides divided by }6} \\
x=\sqrt[4]{{{7}\over{6}}} &\lor& x=-\sqrt[4]{{{7}\over{6}}} \\
&&\phantom{xxx}\blue{\text{both sides taken the }4 \text{-th root}}\\
x=\frac{1}{6}\sqrt[4]{1512} &\lor& x=-\frac{1}{6}\sqrt[4]{1512}\\
&&\phantom{xxx}\blue{\text{simplified}} \end{array}#
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