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}{3}\sqrt{21} \lor x=\frac{1}{3}\sqrt{21}#
#\begin{array}{rcl}3\, x^{2}+2&=& 9 \\
&&\phantom{xxx}\blue{\text{the equation we need to solve}} \\
3\, x^{2}&=&7 \\
&&\phantom{xxx}\blue{\text{both sides minus }2} \\
x^{2} &=& {{7}\over{3}} \\
&&\phantom{xxx}\blue{\text{both sides divided by }3} \\
x=\sqrt[2]{{{7}\over{3}}} &\lor& x=\sqrt[2]{{{7}\over{3}}} \\
&&\phantom{xxx}\blue{\text{both sides taken the }2 \text{th root}}\\
x=\frac{1}{3}\sqrt{21} &\lor& x=\frac{1}{3}\sqrt{21}\\
&&\phantom{xxx}\blue{\text{simplified}} \end{array}#
#\begin{array}{rcl}3\, x^{2}+2&=& 9 \\
&&\phantom{xxx}\blue{\text{the equation we need to solve}} \\
3\, x^{2}&=&7 \\
&&\phantom{xxx}\blue{\text{both sides minus }2} \\
x^{2} &=& {{7}\over{3}} \\
&&\phantom{xxx}\blue{\text{both sides divided by }3} \\
x=\sqrt[2]{{{7}\over{3}}} &\lor& x=\sqrt[2]{{{7}\over{3}}} \\
&&\phantom{xxx}\blue{\text{both sides taken the }2 \text{th root}}\\
x=\frac{1}{3}\sqrt{21} &\lor& x=\frac{1}{3}\sqrt{21}\\
&&\phantom{xxx}\blue{\text{simplified}} \end{array}#
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