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?
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
$\displaystyle \sqrt {3 - \sqrt {x + 5}} = 2$ ........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
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