Thread: find the domain and solve polynomial

Please help me on this.
1. solve x^6+3x^3-(3/4)=0
this is what i've got for this problem
x^3(x^3+3)-(3/4)=0 what should i do next. Quadratic formula???
2.find domain of sqrt((x-3)/(x+2))+1
sorry i dont how to begin for this problem

Originally Posted by Vkenny

Please help me on this.
1. solve x^6+3x^3-(3/4)=0
this is what i've got for this problem
x^3(x^3+3)-(3/4)=0 what should i do next. Quadratic formula???

um, quadratic formula only works for quadratics. you do not have a quadratic in this simplification.

write it this way:

aha! now we have a quadratic! (to see this, replace x^3 with y)

2.find domain of sqrt((x-3)/(x+2))+1
sorry i dont how to begin for this problem

the domain is the set of x-values for which the function is defined. usually the easiest way to find the domain is to find the x-values that don't work, and say the domain is all x-values but those. so, what can x NOT be here?

1.x=+or- 3sqrt((-3+sqrt(12))/2)

2.the domain is (2x-1)/(x+2) greater or equal 0

are these right. thank you for your help

Hello Kenny:

I checked your two solutions for the first exercise by substituting them into the original polynomial and evaluating. I did not get a value of zero from either of them.

(Your positive solution yields 17.4135 and your negative solution yields -0.6953.)

Check your work on solving the quadratic equation, and make sure that you correctly take the cube root of each result.

On the domain exercise, nice try (heh, heh), but your instructor probably won't accept an unsolved inequality as an acceptable answer.

Whenever you report a domain, you need to state a set of numbers because that is what the domain is, a set of numbers.

(You're kinda on the right track.)

The radicand must be non-negative, so you need to solve the following inequality.

(x - 3)/(x + 2) ≥ 0

In other words, the set of numbers for x that satisfy this inequality is the domain of the original function.


If you need more help with either of these exercises, then please continue to post your work and try to say something about why you're stuck.

Cheers,

~ Mark