SOLUTION: Quartic equations of the form {{{Ax^4+Bx^3+Cx^2+Bx+A=0}}}, (A does not equal 0), are reducible to quadratics using the substitution {{{u=x+1/x }}} and grouping terms appropriately.

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: Quartic equations of the form {{{Ax^4+Bx^3+Cx^2+Bx+A=0}}}, (A does not equal 0), are reducible to quadratics using the substitution {{{u=x+1/x }}} and grouping terms appropriately.      Log On


   

Question 1126759: Quartic equations of the form Ax%5E4%2BBx%5E3%2BCx%5E2%2BBx%2BA=0, (A does not equal 0), are reducible to quadratics using the substitution u=x%2B1%2Fx+ and grouping terms appropriately. Solve for x given x%5E4+%2B3x%5E3-8x%5E2+%2B3x%2B1=0.
Answer by greenestamps(13198) About Me  (Show Source):
You can put this solution on YOUR website!


x%5E4+%2B3x%5E3-8x%5E2+%2B3x%2B1=0

Divide the whole equation by x^2. (We can do this because we can see by inspection that x=0 is not a root.)

x%5E2%2B3x-8%2B3%2Fx%2B1%2Fx%5E2+=+0

x%5E2%2B2%2B1%2Fx%5E2 is (x+1/x) squared. Use this to rewrite the expression on the left in the equation as a polynomial with "x+1/x" as the variable; then factor.

%28x%5E2%2B2%2B1%2Fx%5E2%29%2B3x%2B3%2Fx-10+=+0 [group terms as appropriate; note the "-8" is now represented as "2...-10"]

%28x%2B1%2Fx%29%5E2%2B3%28x%2B1%2Fx%29-10+=+0 [this is now a polynomial with "x+1/x" as the variable. Substitute u = x+1/x if it helps you see what is being done]

%28%28x%2B1%2Fx%29%2B5%29%28%28x%2B1%2Fx%29-2%29+=+0

Now multiply each factor by x (to "un-do" the first step above where we divided the whole equation by x^2)

%28x%5E2%2B5x%2B1%29%28x%5E2-2x%2B1%29+=+0

x%5E2%2B5x%2B1+=+0 or x%5E2-2x%2B1+=+0

x+=+%28-5+%2B-+sqrt%2821%29%29%2F2 or x+=+1