Finding the root of a quadratic function

Flkeavy

j(x) = 3x^10 - 33x^5 +1
the answers are x= 1/2 and x= 1 but I dont understand how to get there.

Reply 1

fliscia

Original post by Flkeavy

j(x) = 3x^10 - 33x^5 +1
the answers are x= 1/2 and x= 1 but I dont understand how to get there.

Try making a letter (like y) equal x^5 and work from there

Reply 2

RogerOxon

Original post by Flkeavy

j(x) = 3x^10 - 33x^5 +1
the answers are x= 1/2 and x= 1 but I dont understand how to get there.

No, those aren't the roots. If you substitute them into your function, you don't get 0. x=1/2 is close though.

Is the equation correct?

(edited 6 years ago)

Reply 3

RichardM37

I am curious: how did you end up with a tenth-degree polynomial? It seems a little advanced compared to some of the other work that is discussed in these forums.

In any case, "fliscia" gave some good advice, make a substitution for x^5. Then the equation becomes a second degree quadratic equation:

e.g., u = x^5

The original equation becomes f(u) = 3 u^2 - 33 u + 1
This equation is a quadratic equation in u; you can solve for u by using the quadratic formula. You'd have two roots. Knowing u, you can solve for x.

Alternately, you could solve the original equation numerically. There are many free tools available online. Since the polynomial is of degree 10, there are 10 roots. I quickly ran this problem through a numerical solver: Polynomial Root-finder

Only two of the ten roots are real; the other eight are complex. And the two real ones are not 1 and 1/2.