find the domain

[vkenny]

1.Find the domain: sqrt((x-3)/(x+2))+1
i found the domain: X>0 but my teacher said It's wrong.

2.and solve x^6+3x^3-(3/4)=0
this is what i have for this problem
x^3(x^3+3)-(3/4)=0 so what i should i do next

thank you for helping me.

 

[Loren]

For #1 if x=1 you get the square root of a negative number. Is that ok?

 

[vkenny]

For #1 if x=1 you get the square root of a negative number. Is that ok?

i dont think so, cant have negative number inside the square root.

 

[Deleted member 4993]

1.Find the domain: sqrt((x-3)/(x+2))+1
i found the domain: X>0 but my teacher said It's wrong. At x = 2, what is the value of \(\displaystyle \sqrt{x-3}\)

2.and solve x^6+3x^3-(3/4)=0 <<< substitute u = x[sup:vt2bpdad]3[/sup:vt2bpdad] - then you'll have a quadratic - solve for 'u' - then solve for 'x'
this is what i have for this problem
x^3(x^3+3)-(3/4)=0 so what i should i do next

thank you for helping me.

 

[vkenny]

1. i found the domain is X cant not be 2
is this right.
2. thank you for your help. i know how to do is know. i wished i could show you my answer but i dont know how to type it.

 

[Deleted member 4993]

1. i found the domain is X cant not be 2 << No - can x be 2.99? what about x = 2.99999??
is this right.
2. thank you for your help. i know how to do is know. i wished i could show you my answer but i dont know how to type it.