Functions: Higher degree polynomials
Solving higher degree polynomials with the quadratic formula
Some equations with polynomials can be solved with the quadratic formula. For that, we use substitution.
Procedure We solve an equation with polynomials in #x# with the quadratic formula. 
Example #2x^4+3x^22=0# 

Step 1  Write the equation in the form #a \blue x^{\blue n \cdot 2}+b \blue{x^n} +c=0#.  #2\blue{x}^{\blue2 \cdot 2}+3\blue{x^2}2=0# 
Step 2  Substitute #\blue{x^n}=\green u#.  #2\green u^2+3\green u2=0# 
Step 3  Solve the obtained quadratic equation in #\green u# with the quadratic formula.  #\green u=2 \lor \green u =\tfrac{1}{2}# 
Step 4  Substitute #\green u =\blue{x^n}# in the found solution(s).  #\blue{x^2}=2 \lor \blue{x^2}=\tfrac{1}{2}# 
Step 5  Determine the solutions in #x# from the equations obtained in step 4.  #x=\tfrac{1}{\sqrt{2}} \lor x=\tfrac{1}{\sqrt{2}}# 
#x=\sqrt[5]{{{3}\over{5}}} \lor x=\sqrt[5]{{{9}\over{5}}} #
Step 1  We write the equation in the form: \[25 x^{2 \cdot 5}60 x^{5}+27=0\] 
Step 2  We substitute #x^5=u#. This gives: \[25 u^260 u+27=0\] 
Step 3  We solve the obtained equation in #u# by means of the quadratic formula. The discriminant is equal to: \[\begin{array}{rcl}D&=&b^24ac \\ &&\phantom{xxx}\blue{\text{formula for the discriminant}}\\ &=& \left(60\right)^24 \cdot 25 \cdot 27 \\ &&\phantom{xxx}\blue{\text{formula entered}}\\ &=& 900 \\ &&\phantom{xxx}\blue{\text{calculated}}\end{array}\] Since the discriminant is positive, there are two solutions. These are: \[\begin{array}{rcl}u=\frac{b\sqrt{D}}{2a} &\lor& u=\frac{b+\sqrt{D}}{2a} \\ &&\phantom{xxx}\blue{\text{formula for the solutions}}\\ u=\frac{{60}\sqrt{900}}{2 \cdot 25} &\lor& u=\frac{{60}+\sqrt{900}}{2 \cdot 25}\\ &&\phantom{xxx}\blue{\text{formula entered}}\\ u={{3}\over{5}} &\lor& u={{9}\over{5}} \\ &&\phantom{xxx}\blue{\text{calculated}}\end{array}\] 
Step 4  Now we substitute #u=x^{5}# in the found solutions, this gives us \[x^{5}={{3}\over{5}} \lor x^{5}={{9}\over{5}}\] 
Step 5  Finally we solve these equations by taking the root. This gives us the solutions to the original equation: \[x=\sqrt[5]{{{3}\over{5}}} \lor x=\sqrt[5]{{{9}\over{5}}}\] 
Unlock full access
Teacher access
Request a demo account. We will help you get started with our digital learning environment.
Student access
Is your university not a partner?
Get access to our courses via Pass Your Math independent of your university. See pricing and more.
Or visit omptest.org if jou are taking an OMPT exam.
Or visit omptest.org if jou are taking an OMPT exam.