Functions: Power functions and root functions
Solving root equations with substitution
Just as with polynomials there exist root equations that can be solved by using the techniques for solving quadratic equations.
Solving root equations by substitution
Procedure We solve the root equation for #x# by means of substitution. 
Example #x^35 x \sqrt{x}+6=0# 

Step 1 
Write the equation as #a\blue x^{\blue n \cdot 2}+b\blue x^{\blue n}+c=0#, in which #a#, #b# and #c# are numbers and #\blue n# a fraction. 
#\blue{x}^{\blue{\frac{3}{2}} \cdot 2}5\blue x^{\blue{\frac{3}{2}}}+6=0# 
Step 2  Substitute #\blue{x^n}=\green u#. 
#\green u^25u+6=0# 
Step 3  Solve the obtained equation for #u#. 
#\green u=3 \lor \green u=2# 
Step 4  Substitute #\green u=\blue{x^n}# in the solutions. 
#\blue{x^{\frac{3}{2}}}=3 \lor \blue{x^{\frac{3}{2}}}=2# 
Step 5  Solve the obtained equations. 
#x=\sqrt[3]{9} \lor x=\sqrt[3]{4}# 
Step 6  Check the solutions in the original equation. 
#(\sqrt[3]{9})^35 \cdot \sqrt[3]{9} \sqrt{\sqrt[3]{9}}+6=0# #(\sqrt[3]{4})^35 \cdot \sqrt[3]{4} \sqrt{\sqrt[3]{4}}+6=0# Hence, the solution to the equation is #x=\sqrt[3]{9} \lor x=\sqrt[3]{4}#. 
Solve the equation #x8\cdot \sqrt{x}+15=0# with unknown #x#, using substitution. Provide an answer of the form #x=x_1\vee x=x_2# if there are two solutions, of the form #x=x_1# if there is a single solution, and write #none# if there are no solutions.
#x=9\lor x=25#
Step 1  We write the equation as: \[x^{\frac{1}{2} \cdot 2}8x^{\frac{1}{2}}+15\] 
Step 2  Now we substitute #x^{\frac{1}{2}}=u#. This gives us: \[u^28\cdot u+15=0\] 
Step 3  We solve the found equation by means of factorization. This is done like this: \[\begin{array}{rcl}u^28\cdot u+15 &=& 0\\&&\phantom{xxx}\blue{\text{the original equation}}\\ \left(u3\right)\left(u5\right)&=&0\\&&\phantom{xxx}\blue{\text{factorized}}\\ u3=0&\lor&u5=0\\&&\phantom{xxx}\blue{A\cdot B=0 \Leftrightarrow A=0 \lor B=0}\\ u=3&\lor& u=5\\&&\phantom{xxx}\blue{\text{constant terms to the right hand side}} \end{array}\] 
Step 4  Now we substitute #u=x^{\frac{1}{2}}# into the solutions. This gives: \[x^{\frac{1}{2}}=3 \lor x^{\frac{1}{2}}=5\] 
Step 5  To solve these equations, we substitute both sides to the power #2#. This gives the possible solutions: \[x=3^{2} \lor x=5^{2}\] This can be simplified to: \[x=9 \lor x=25\] 
Step 6  Now we check both solutions. First #x=9#, this gives: \[98\cdot \sqrt{9}+15=0\] Next #x=25#, this gives: \[258\cdot \sqrt{25}+15=0\] Hence, both solutions are correct and the solution of the equation is: \[x=9\lor x=25\] 
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