Pre-Calculus Help!

**rsultana** · September 30th 2007, 03:14 PM

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?

**polymerase** · September 30th 2007, 03:16 PM

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?

$ (x^2-2)(x^2-1) $

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

**Jhevon** · September 30th 2007, 03:28 PM

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

**rsultana** · September 30th 2007, 03:31 PM

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

**rsultana** · September 30th 2007, 03:32 PM

Originally Posted by polymerase

$ (x^2-2)(x^2-1) $

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

can you please explain this to me. oh never mind i got it!

**Jhevon** · September 30th 2007, 03:36 PM

Originally Posted by rsultana

can you please explain this to me.

he rewrote $x^4 - 3x^2 + 2$ as $\left( x^2 \right)^2 - 3 \left( x^2 \right) + 2$ , which is quadratic in $x^2$ , (you can see this if you replaced $x^2$ with $y$ for instance, you would get, $y^2 - 3y + 2$ ).

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

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

both of these last terms are the difference of two squares

**Jhevon** · September 30th 2007, 03:38 PM

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

$\sqrt {3 - \sqrt {x + 5}} = 2$ ........square both sides

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

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

$\Rightarrow x + 5 = 1$

now continue

remember to check your solutions at the end

**rsultana** · September 30th 2007, 05:26 PM

Originally Posted by Jhevon

$\sqrt {3 - \sqrt {x + 5}} = 2$ ........square both sides

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

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

$\Rightarrow x + 5 = 1$

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?
please explain this problem!

**topsquark** · September 30th 2007, 05:42 PM

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?
please explain this problem!

Cosine? Where did that come from?

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

-Dan

**Jhevon** · September 30th 2007, 05:45 PM

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?
please explain this problem!

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

**Jhevon** · September 30th 2007, 05:46 PM

Originally Posted by topsquark

Cosine? Where did that come from?

hehe, i think the poster meant "'cause"

**rsultana** · September 30th 2007, 05:46 PM

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 $\sqrt{2} = 2$ , 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

**topsquark** · September 30th 2007, 07:14 PM

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

**rsultana** · September 30th 2007, 07:15 PM

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

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