1. ## I need help!!!

Hi ppl I really need help in some exercise!!!

1)Nth root seventeen ___ Nth root 6 + Nth root 3
2)Nth root seven + Nth root five ____ Nth root 2 + Nth root twelve

3) x+4= Nth root (inside of it: 3x^2+5x-4)

4) (x^2+3x)^2 -14(x^2 +3x) +40 = 0
5) (x+6/x)^2 -12(x+6/x)+35=0

plz help me!!!!!!! i will be grateful to the one who show mw how i need solve atleast one of those...

p.s
i solve(with your help,thank you!!!) all the exercise except number 3.help plz:]

2. Originally Posted by katia15
4) (x^2+3x)^2 -14(x^2 +3x) +40 = 0
Put y=x^2+3x, then:

y^2-14y+40=0

which has roots y=4 and y=10, so now solve:

x^2+3x=4

(roots are by inspection x=-4, x=1)

and

x^2+3x=10

(roots are by inspection x=-5 and x=2)

to find all the x's that satisfy the original equation.

RonL

3. Originally Posted by katia15
5) (x+6/x)^2 -12(x+6/x)+35=0
Same trick as for (4), put y=x+6/x, to get y^2-12y+35=0, and take it from there.

RonL

4. Hello, katia15!

It's about comparison: less than or greater than.

$\displaystyle 1)\; \sqrt[n]{17}\;\; ^{>}_{<} \;\;\sqrt[n]{6}+ \sqrt[n]{3}$
I assume that the inequality holds for any positive interger $\displaystyle n$.

Let's test it for $\displaystyle n=2$.

We have: .$\displaystyle \sqrt{17} \;\;^{>}_{<}\;\;\sqrt{6} + \sqrt{3}$

Square both sides: .$\displaystyle \left(\sqrt{17}\right)^2\;\;^{>}_{<}\;\;\left(\sqr t{6} + \sqrt{3}\right)^2$

And we have: .$\displaystyle 17 \;\;^{>}_{<}\;\;6 + 2\sqrt{18} + 3\quad\Rightarrow\quad8\;\;^{>}_{<}\;\;2\sqrt{18}\ quad\Rightarrow\quad 4\;\;^{>}_{<}\;\;\sqrt{18}$

Square both sides: .$\displaystyle (4)^2\;\;^{>}_{<}\;\;\left(\sqrt{18}\right)^2\quad \Rightarrow\quad 16\;\;^{>}_{<}\;\;18$

Since $\displaystyle 16 < 18$, the original inequality is "less than".

Therefore: .$\displaystyle \sqrt[n]{17}\;\;{\color{red}<}\;\;\sqrt[n]{6} + \sqrt[n]{3}$

5. ## thanks:)

hi i wanted to sat thank you for CaptainBlack and soroban.
you helpes me a lot