document.write( "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 $\"x%5E4+%2B3x%5E3-8x%5E2+%2B3x%2B1=0\"$ . \n" ); document.write( "

Algebra.Com's Answer #743098 by greenestamps(13200) $\"\"$ $\"About$

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\n" ); document.write( " $\"x%5E4+%2B3x%5E3-8x%5E2+%2B3x%2B1=0\"$

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

\n" ); document.write( " $\"x%5E2%2B3x-8%2B3%2Fx%2B1%2Fx%5E2+=+0\"$

\n" ); document.write( " $\"x%5E2%2B2%2B1%2Fx%5E2\"$ 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.

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

\n" ); document.write( " $\"%28x%2B1%2Fx%29%5E2%2B3%28x%2B1%2Fx%29-10+=+0\"$ [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]

\n" ); document.write( " $\"%28%28x%2B1%2Fx%29%2B5%29%28%28x%2B1%2Fx%29-2%29+=+0\"$

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

\n" ); document.write( " $\"%28x%5E2%2B5x%2B1%29%28x%5E2-2x%2B1%29+=+0\"$

\n" ); document.write( " $\"x%5E2%2B5x%2B1+=+0\"$ or $\"x%5E2-2x%2B1+=+0\"$

\n" ); document.write( " $\"x+=+%28-5+%2B-+sqrt%2821%29%29%2F2\"$ or $\"x+=+1\"$ \n" ); document.write( "

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