# Thread: finding real solutions of these examples

1. ## finding real solutions of these examples

lets try this latex thing out...

$\displaystyle 3x^4-2x^2-1=0$

I know I need to move the one to the other side, then I think factor our what is common, in this case.. it would be a x^2 right? so I get to

$\displaystyle x^2(3x^2-2)=1$

I get stumped after that.

Another one I have is
$\displaystyle x^6+7x^3-8=0$

2. Originally Posted by Geomatt
lets try this latex thing out...

$\displaystyle 3x^4-2x^2-1=0$

I know I need to move the one to the other side, then I think factor our what is common, in this case.. it would be a x^2 right? so I get to

$\displaystyle x^2(3x^2-2)=1$

I get stumped after that.

Another one I have is
$\displaystyle x^6+7x^3-8=0$
The first is quadratic in $\displaystyle x^2$, the second is quadratic in $\displaystyle x^3$.

Note that:

$\displaystyle 3x^4 - 2x^2 - 1 = 3 \left( x^2 \right)^2 - 2 \left( x^2 \right) - 1 = 0$

similarly,

$\displaystyle x^6 + 7x^3 - 8 = \left( x^3 \right)^2 + 7 \left( x^3 \right) - 8 = 0$

can you take it from here?

3. Originally Posted by Jhevon
Note that:

$\displaystyle 3x^4 - 2x^2 - 1 = 3 \left( x^2 \right)^2 - 2 \left( x^2 \right) - 1 = 0$

similarly,

$\displaystyle x^6 + 7x^3 - 8 = \left( x^3 \right)^2 + 7 \left( x^3 \right) - 8$

can you take it from here?
lets see, so it would be $\displaystyle 3x-2x^2-1=0$ which I could then rearrange and solve with the quad. formula?

im not sure about the second example.

Im sorry if the answer is obvious, math has always been tough to me.

4. Originally Posted by Geomatt
lets see, so it would be $\displaystyle 3x-2x^2-1=0$ which I could then rearrange and solve with the quad. formula?

im not sure about the second example.

Im sorry if the answer is obvious, math has always been tough to me.
for the first:

how would you solve $\displaystyle 3y^2 - 2y - 1 = 0$? what is the similarity between this last equation and the one i gave you for the first?

5. Originally Posted by Jhevon
for the first:

how would you solve $\displaystyle 3y^2 - 2y - 1 = 0$? what is the similarity between this last equation and the one i gave you for the first?
ok so
the first example
$\displaystyle 3x-2x^2-1=0$
$\displaystyle 2x^2-3x+1=0$
$\displaystyle (2x-1)(x-1)$
x= -1/2, x=1

I checked my answers and im supposed to get x=1, -1, i messed up somewhere.

6. Originally Posted by Geomatt
ok so
the first example
$\displaystyle 3x-2x^2-1=0$
$\displaystyle 2x^2-3x+1=0$
$\displaystyle (2x-1)(x-1)$
x= -1/2, x=1

I checked my answers and im supposed to get x=1, -1, i messed up somewhere.
this is wrong because you have the wrong question. it is $\displaystyle 3x^4 - 2x^2 - 1 = 0$ you forgot the fourth power

i'll do the first, you do the second

$\displaystyle 3x^4 - 2x^2 - 1 = 0$ ......rewrite

$\displaystyle \Rightarrow 3 \left( x^2 \right)^2 - 2 \left( x^2 \right) - 1 = 0$ ........now factor as if the $\displaystyle x^2$ was just a variable by itself, like $\displaystyle y$ or something

$\displaystyle \Rightarrow \left(3x^2 + 1 \right) \left( x^2 - 1 \right) = 0$

$\displaystyle \Rightarrow x^2 = - \frac {1}{3} \mbox { or } x^2 = 1$

$\displaystyle \Rightarrow x = \sqrt { \frac {1}{3}}i \mbox { or } x = \pm 1$

you don't want the complex roots, so just take the $\displaystyle \pm 1$ answer, that's the real roots solution

7. Originally Posted by Jhevon
$\displaystyle \Rightarrow x^2 = - \frac {1}{3} \mbox { or } x^2 = 1$

$\displaystyle \Rightarrow x = \sqrt { \frac {1}{3}}i \mbox { or } x = \pm 1$

$\displaystyle \Rightarrow x = \pm \sqrt { \frac {1}{3}}~i \mbox { or } x = \pm 1$

RonL

8. Originally Posted by CaptainBlack
$\displaystyle \Rightarrow x = \pm \sqrt { \frac {1}{3}}~i \mbox { or } x = \pm 1$

RonL
yeah, that's what i meant