1. ## Help with reducing quadratic equation to final terms

$x^4 - 6x^2 + 8 = 0
$

$x^2=t$
$x^4=t^2$

a= 1, b = 6, c = 8

$\Rightarrow x = \frac {6 \pm \sqrt {6^2 - 4(1)(8)}}{2(1)}$

$\Rightarrow x = \frac {6 \pm \sqrt {4}}{2}$

$\Rightarrow x = \frac {3 \pm \sqrt {2}}{}$

How do I reduce this further to get the final answer or is my answer correct? I always get confused. Thanks in advance.

2. Originally Posted by lilrhino
$x^4 - 6x^2 + 8 = 0
$

$x^2=t$
$x^4=t^2$

a= 1, b = 6, c = 8

$\Rightarrow x = \frac {6 \pm \sqrt {6^2 - 4(1)(8)}}{2(1)}$

$\Rightarrow x = \frac {6 \pm \sqrt {4}}{2}$

$\Rightarrow x = \frac {3} \pm \frac { \sqrt {2}}{2}$

How do I reduce this further to get the final answer. I always get confused. Thanks in advance.
ok, so this is wrong, since b = -6 not 6. but we don't need the quadratic formula here

you replaced $x^2$ with $t$, so you should have

$t^2 - 6t + 8 = 0$ .........now factor

$\Rightarrow (t - 4)(t - 2) = 0$

$\Rightarrow t = 4 \mbox { or } t = 2$

But $t = x^2$

$\Rightarrow x^2 = 4 \mbox { or } x^2 = 2$

$\Rightarrow x = \pm 2 \mbox { or } x = \pm \sqrt {2}$

3. Hello, lilrhino!

You're making it unnecessarily complicated . . . and wrong.

$x^4 - 6x^2 + 8 \:= \:0$
Factor: . $(x^2 - 4)(x^2 - 2)\:=\:0$

And we have two equations to solve:

. . $x^2 - 4\:=\:0\quad\Rightarrow\quad x^2 \:=\:4\quad\Rightarrow\quad x \:=\: \pm2$

. . $x^2 - 2\:=\:0\quad\Rightarrow\quad x^2\:=\:2\quad\Rightarrow\quad x\:=\:\pm\sqrt{2}$

4. Originally Posted by Jhevon
ok, so this is wrong, since b = -6 not 6. but we don't need the quadratic formula here

you replaced $x^2$ with $t$, so you should have

$t^2 - 6t + 8 = 0$ .........now factor

$\Rightarrow (t - 4)(t - 2) = 0$

$\Rightarrow t = 4 \mbox { or } t = 2$

But $t = x^2$

$\Rightarrow x^2 = 4 \mbox { or } x^2 = 2$

$\Rightarrow x = \pm 2 \mbox { or } x = \pm \sqrt {2}$
Thanks Jhevon, I have to watch my signs. I was looking at an example in the book, so I thought I had to use the quadratic formula even though it could be factored as you demonstrated. Thanks again!

5. Originally Posted by lilrhino
Thanks Jhevon, I have to watch my signs. I was looking at an example in the book, so I thought I had to use the quadratic formula even though it could be factored as you demonstrated. Thanks again!
yeah, the quadratic formula can be overkill at times. However, i do agree with Soroban that we are making it more complicated than it should be. Generally it helps students to visualize what to do by replacing a squared term with a single variable, but i believe that's wasting writing space. ordinarily, i'd just factor as Soroban did and cut out the middle man, but i didn't really want to shock you. you'd be surprised at some of the things students are shocked with

6. Originally Posted by Soroban
Hello, lilrhino!

You're making it unnecessarily complicated . . . and wrong.

Factor: . $(x^2 - 4)(x^2 - 2)\:=\:0$

And we have two equations to solve:

. . $x^2 - 4\:=\:0\quad\Rightarrow\quad x^2 \:=\:4\quad\Rightarrow\quad x \:=\: \pm2$

. . $x^2 - 2\:=\:0\quad\Rightarrow\quad x^2\:=\:2\quad\Rightarrow\quad x\:=\:\pm\sqrt{2}$
I do that often unfortunately. Math is not my strongest subject, so I sometimes make things more complicated that necessary. Thanks for your response.

7. Originally Posted by Jhevon
yeah, the quadratic formula can be overkill at times. However, i do agree with Soroban that we are making it more complicated than it should be. Generally it helps students to visualize what to do by replacing a squared term with a single variable, but i believe that's wasting writing space. ordinarily, i'd just factor as Soroban did and cut out the middle man, but i didn't really want to shock you. you'd be surprised at some of the things students are shocked with
I'm not easily shocked, I may spend more time thinking about it though . It's easier the way Soroban demonstrated it actually, so I'm not shocked. Thanks...