I've had a go at this but I can't seem to get anywhere with it.

2x^4 + x^3 - 6x^2 + x + 2 = 0

Use the substitution p = x + 1/x to get a quadratic in p.

I've tried just substituting in x = p - 1/x but the x's don't cancel. How should I approach this problem?

2. ## Re: Disguised quadratic equation

Well, x+2 sure looks good as a factor....so try long division...

3. ## Re: Disguised quadratic equation

Originally Posted by kinhew93
I've had a go at this but I can't seem to get anywhere with it.

2x^4 + x^3 - 6x^2 + x + 2 = 0

Use the substitution p = x + 1/x to get a quadratic in p.

I've tried just substituting in x = p - 1/x but the x's don't cancel. How should I approach this problem?

1. If the coefficients of the equation are symmetric to the central summand then divide through by $x^2$. You'll get:

$2x^2 + x - 6 + \frac1x + \frac2{x^2}=0$

2. If $p = x + \frac1x$ then $p^2 = x^2 + 2 + \frac1{x^2} ~\implies~ p^2-2 = x^2 + \frac1{x^2}$

3. Group the equation of #1 to:

$2 \left(x^2+\frac1{x^2}\right) + \left(x+\frac1x\right)-6=0$

Now replace the terms in brackets by the terms of #2:

$2(p^2-2)+p-6=0~\implies~2p^2+p-10=0$

which yields: $p = 2~\vee~p=-\frac52$

4. Now replace p in the equation of #2 and solve for x:

$2 = x+\frac1x~\vee~-\frac52 = x+\frac1x$

$x^2-2x+1 = 0~\vee~x^2+\frac52 x +1 = 0$

$x=1~\vee~x=-\frac12\vee x=-2$

4. ## Re: Disguised quadratic equation

Thanks Wilmer I'm sure that would work but I'm looking to solve it using the substitution given.