Solving higher degree polynomials with the quadratic formula

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^2-2=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 u-2=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}}#

Solve #x# in the equation
\[2\cdot x^{10}+15\cdot x^5-27=0\tiny\]
Give your answer in the form #x=x_1\lor x=x_2 \lor \ldots \lor x=x_n#, in which #x_1#, #x_2#, #\ldots#, #x_n# are numbers and #n# the number of solutions.

#x=\sqrt[5]{-9} \lor x=\sqrt[5]{{{3}\over{2}}} #

Step 1	We write the equation in the form: \[2 x^{2 \cdot 5}+15 x^{5}-27=0\]
Step 2	We substitute #x^5=u#. This gives: \[2 u^2+15 u-27=0\]
Step 3	We solve the obtained equation in #u# by means of the quadratic formula. The discriminant is equal to: \[\begin{array}{rcl}D&=&b^2-4ac \\ &&\phantom{xxx}\blue{\text{formula for the discriminant}}\\ &=& \left(15\right)^2-4 \cdot 2 \cdot -27 \\ &&\phantom{xxx}\blue{\text{formula entered}}\\ &=& 441 \\ &&\phantom{xxx}\blue{\text{calculated}}\end{array}\] Since the discriminant is positive, there are two solutions. These are: \[\begin{array}{rcl}u=\frac{-b-\sqrt{D}}{2a} &\lor& u=\frac{-b+\sqrt{D}}{2a} \\ &&\phantom{xxx}\blue{\text{formula for the solutions}}\\ u=\frac{-{15}-\sqrt{441}}{2 \cdot 2} &\lor& u=\frac{-{15}+\sqrt{441}}{2 \cdot 2}\\ &&\phantom{xxx}\blue{\text{formula entered}}\\ u=-9 &\lor& u={{3}\over{2}} \\ &&\phantom{xxx}\blue{\text{calculated}}\end{array}\]
Step 4	Now we substitute #u=x^{5}# in the found solutions, this gives us \[x^{5}=-9 \lor x^{5}={{3}\over{2}}\]
Step 5	Finally we solve these equations by taking the root. This gives us the solutions to the original equation: \[x=\sqrt[5]{-9} \lor x=\sqrt[5]{{{3}\over{2}}}\]

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