Lesson Solving polynomial equations of high degree by introducing a new variable
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<H2>Solving polynomial equations of high degree by introducing a new variable</H2> <H3>Problem 1</H3>Solve an equation of degree 4 {{{x^4}}} - {{{13x^2}}} + {{{36}}} = {{{0}}}. <B>Solution</B> Introduce a new variable {{{y}}} = {{{x^2}}}. Then you get a quadratic equation {{{y^2}}} - {{{13y}}} + {{{36}}} = {{{0}}}. Solve it. You can use the quadratic formula, or the Viete's theorem, or factoring. The roots of this equation are y=4 and y=9. Correspondingly, the roots of the original equation are x = +/-2 and x = +/-3. <H3>Problem 2</H3>Solve an equation of degree 6 {{{x^6}}} + {{{3x^3}}} + {{{2}}} = {{{0}}}. <B>Solution</B> Introduce a new variable {{{y}}} = {{{x^3}}}. Then you get a quadratic equation {{{y^2}}} + {{{3y}}} + {{{2}}} = {{{0}}}. Solve it. You can use the quadratic formula, or the Viete's theorem, or factoring. The roots of this equation are y=-1 and y=-2. Correspondingly, the roots of the original equation are {{{root(3,-1)}}} and {{{root(3,-2)}}}. {{{root(3,-1)}}} has three values: First one is the real number -1, and two others are complex numbers {{{(1/2) + (sqrt(3)/2)*i}}} and {{{(1/2) - (sqrt(3)/2)*i}}}. {{{root(3,-2)}}} has also three values. First one is the real number -{{{root(3,2)}}} , and two others are complex numbers {{{root(3,2)*((1/2) + (sqrt(3)/2)*i)}}} and {{{root(3,2)*((1/2) - (sqrt(3)/2)*i)}}}. <H3>Problem 3</H3>f(x) = x^3 - 2 and g(x) = x^2 - 5x. Solve equation (gof)(x) = 6. <B>Solution</B> Notice that the left side of the equation is the <U>composition</U> of polynomials gof. <pre> So, they want you solve this equation {{{(x^3-2)^2 -5*(x^3-2)}}} = 6, Which is, obviously, polynomial equation of degree 6. Introduce new variable y = x^3 - 2. (In other words, y = f(x) ). Then the given equation takes the form y^2 - 5y = 6 or y^2 - 5y - 6 = 0. (1) Factor left side (y-6)*(y+1) = 0, which gives the roots y= 6 and y= -1. If y= 6, then x^3 - 2 = 6, x^3 = 6 + 2 = 8, which implies x = {{{root(3,8)}}} = 2. If y= -1, then x^3 - 2 = -1, x^3 = -1 + 2 = 1, which implies x = {{{root(3,1)}}} = 1. So, the real roots of the equation (1) are the values 1 and/or 2. If you want to get all complex roots of equation (1), you should obtain complex roots of equations x^3 = 8 and x^3 = 1. They are x = {{{2*cis(2pi/3)}}} and {{{2*cis(4pi/3)}}} for equation x^3 = 8, and x = {{{cis(2pi/3)}}} and {{{cis(4pi/3)}}} for equation x^3 = 1. Thus the full list of the solutions to the given equation (1) 1, {{{cis(2pi/3)}}}, {{{cis(4pi/3)}}}, 2, {{{2*cis(2pi/3)}}} and {{{2*cis(4pi/3)}}}. </pre> <B><U>Introducing new variable is the standard method for solving equations like these</U></B>. Surely, in those cases when it is applicable. My other closely related lessons in this site are - <A HREF=http://www.algebra.com/algebra/homework/Polynomials-and-rational-expressions/81-Solving-polynomial-equations-of-high-degree-by-factoring.lesson>Solving polynomial equations of high degree by factoring</A> - <A HREF=https://www.algebra.com/algebra/homework/Polynomials-and-rational-expressions/Advanced-factoring.lesson>Advanced factoring</A> - <A HREF=https://www.algebra.com/algebra/homework/Polynomials-and-rational-expressions/Upper-level-miracle-factoring.lesson>Upper_level_miracle_factoring</A> - <A HREF=https://www.algebra.com/algebra/homework/Polynomials-and-rational-expressions/Solving-palindromic-equations-of--the-degree-4.lesson>Solving palindromic equations of the degree 4</A> - <A HREF=https://www.algebra.com/algebra/homework/Polynomials-and-rational-expressions/OVERVIEW-of-lessons-on-solving-polynomial-equations-of-high-degree-%28Algebra-1%29.lesson>OVERVIEW of lessons on solving polynomial equations of high degree</A> Use this file/link <A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson>ALGEBRA-I - YOUR ONLINE TEXTBOOK</A> to navigate over all topics and lessons of the online textbook ALGEBRA-I.
