1. Pre-Calculus Help!

Need Help with :
finding real solutions.
x4-3x2+2=0
the numbers after the x's are the powers.

also if i need to type in square roots how do i do that?

2. Originally Posted by rsultana
Need Help with :
finding real solutions.
x4-3x2+2=0
the numbers after the x's are the powers.

also if i need to type in square roots how do i do that?
$\displaystyle (x^2-2)(x^2-1)$

$\displaystyle x=+\sqrt{2}, -\sqrt{2},1,-1$

3. Originally Posted by rsultana
Need Help with :
finding real solutions.
x4-3x2+2=0
the numbers after the x's are the powers.

also if i need to type in square roots how do i do that?
ah, i see! you typed the question wrong last time, had me scratching my head

4. Originally Posted by Jhevon
ah, i see! you typed the question wrong last time, had me scratching my head
ooops i am so sorry...hey thanks for the help
Finding real solutions:
sqrt(3-sqrt(x+5))=2 <<<<< i dont know if i am writing this correct. let me know if you understand

5. Originally Posted by polymerase
$\displaystyle (x^2-2)(x^2-1)$

$\displaystyle x=+\sqrt{2}, -\sqrt{2},1,-1$
can you please explain this to me. oh never mind i got it!

6. Originally Posted by rsultana
can you please explain this to me.
he rewrote $\displaystyle x^4 - 3x^2 + 2$ as $\displaystyle \left( x^2 \right)^2 - 3 \left( x^2 \right) + 2$, which is quadratic in $\displaystyle x^2$, (you can see this if you replaced $\displaystyle x^2$ with $\displaystyle y$ for instance, you would get, $\displaystyle y^2 - 3y + 2$).

so he just factored it as if it were a quadratic:

you would get, $\displaystyle (y - 2)(y - 1) = \left( x^2 - 2 \right) \left( x^2 - 1 \right)$

both of these last terms are the difference of two squares

7. Originally Posted by rsultana
ooops i am so sorry...hey thanks for the help
Finding real solutions:
sqrt(3-sqrt(x+5))=2 <<<<< i dont know if i am writing this correct. let me know if you understand
$\displaystyle \sqrt {3 - \sqrt {x + 5}} = 2$ ........square both sides

$\displaystyle \Rightarrow 3 - \sqrt{x + 5} = 4$

$\displaystyle \Rightarrow -\sqrt{x + 5} = 1$ ..........sqaure both sides again

$\displaystyle \Rightarrow x + 5 = 1$

now continue

remember to check your solutions at the end

8. Originally Posted by Jhevon
$\displaystyle \sqrt {3 - \sqrt {x + 5}} = 2$ ........square both sides

$\displaystyle \Rightarrow 3 - \sqrt{x + 5} = 4$

$\displaystyle \Rightarrow -\sqrt{x + 5} = 1$ ..........sqaure both sides again

$\displaystyle \Rightarrow x + 5 = 1$

now continue

remember to check your solutions at the end
my solution manual says no real solution...why?
oooooo...i got it...cos -.3166 is not equal to 2...is this correct?

9. Originally Posted by rsultana
my solution manual says no real solution...why?
oooooo...i got it...cos -.3166 is not equal to 2...is this correct?
Cosine? Where did that come from?

No, it's because when you substitute x = -4 into the original equation, the equation reads that $\displaystyle \sqrt{2} = 2$, which is ridiculous.

-Dan

10. Originally Posted by rsultana
my solution manual says no real solution...why?
oooooo...i got it...cos -.3166 is not equal to 2...is this correct?
this is why i said check your solution at the end. sometimes squaring both sides introduces erroneous solutions. if plugging in the answers you got makes no sense, or don't work, then either there's no solution, or you made a mistake

11. Originally Posted by topsquark
Cosine? Where did that come from?
hehe, i think the poster meant "'cause"

12. Originally Posted by topsquark
Cosine? Where did that come from?

No, it's because when you substitute x = -4 into the original equation, the equation reads that $\displaystyle \sqrt{2} = 2$, which is ridiculous.

-Dan
oops its not cosine it was supposed to be "because".....

another question:
inequality involving quotient: using interval notation, and sketch the solution
2x-3/x+1 ≤1

13. Originally Posted by rsultana
oops its not cosine it was supposed to be "because".....

another question:
inequality involving quotient: using interval notation, and sketch the solution
2x-3/x+1 ≤1
First, new questions should be placed in new threads.

BUT
I already answered this in your other thread. If you cannot understand my solution that thread is the one you should make a mention of it. That's only polite.

AND
Please don't double post. See rule #1 here.

-Dan

14. Originally Posted by topsquark
First, new questions should be placed in new threads.

BUT
I already answered this in your other thread. If you cannot understand my solution that thread is the one you should make a mention of it. That's only polite.

AND
Please don't double post. See rule #1 here.

-Dan
yes sir...i got it...thanks