SOLUTION: The equation {{{x^4 -6x^3 +7x^2 +6x-3=0}}} has irrational roots. Find them by expressing the equation in the form {{{(x^2 +ax)^2 +b(x^2 +ax) +c=0}}}.

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: The equation {{{x^4 -6x^3 +7x^2 +6x-3=0}}} has irrational roots. Find them by expressing the equation in the form {{{(x^2 +ax)^2 +b(x^2 +ax) +c=0}}}.      Log On


   

Question 1126754: The equation x%5E4+-6x%5E3+%2B7x%5E2+%2B6x-3=0 has irrational roots. Find them by expressing the equation in the form %28x%5E2+%2Bax%29%5E2+%2Bb%28x%5E2+%2Bax%29+%2Bc=0.
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


We want to put the given 4th degree polynomial equation in the following form:

%28x%5E2+%2Bax%29%5E2+%2Bb%28x%5E2+%2Bax%29+%2Bc=0

Expand this form and combine like terms:

%28x%5E4%2B2ax%5E3%2Ba%5E2x%5E2%29%2B%28bx%5E2%2Babx%29+%2B+c+=+0

x%5E4%2B%282a%29x%5E3%2B%28a%5E2%2Bb%29x%5E2%2B%28ab%29x%2Bc+=+0

We want this to be equal to

x%5E4+-6x%5E3+%2B7x%5E2+%2B6x-3=0

Solve for a, b, and c by equating coefficients.

x^4: the coefficients are both 1

x^3: 2a+=+-6 --> a = -3

x^2: a%5E2%2Bb+=+%28-3%29%5E2%2Bb+=+9%2Bb+=+7 --> b = -2

x^1: ab+=+6 (we already know that; if the product of our a and b were NOT 6, then the roots could not be found by this method)

x^0: c = -3

The equation in the new form is then

%28x%5E2-3x%29%5E2-2%28x%5E2-3x%29-3+=+0

This is a quadratic equation with "x^2-3x" as the "variable". Factor and solve.

%28%28x%5E2-3x%29-3%29%28%28x%5E2-3x%29%2B1%29+=+0

x%5E2-3x-3+=+0 or x%5E2-3x%2B1+=+0

Neither of these quadratic expressions factors, so we get two pairs of irrational roots.

x+=+%283+%2B-+sqrt%2821%29%29%2F2 and x+=+%283+%2B-+sqrt%285%29%29%2F2