# Math Help - Quadratic in form

I know how to factor this $t^2-16=0$ $(t-4)(t+4)$ but today I am asked to find the real solutions for equations that are quadratic in form....

$t^4-16$....?!

This is how I worked it out, for some reason I think (know) I did something wrong...

$t^4-16=0$

= $(t^2-4)(t^2+4)$

= $(t-2)(t+2)(t+2)(t+2)$

Now should this be the answer $(t-2)(t+2)(t+2)(t+2)$ or should it be this {-2,2}. By solving this I might be able to solve this one $x^4-5x^2+4=0$

Thanks once again guys!!

2. Originally Posted by Morzilla
I know how to factor this $t^2-16=0$ $(t-4)(t+4)$ but today I am asked to find the real solutions for equations that are quadratic in form....

$t^4-16$....?!

This is how I worked it out, for some reason I think (know) I did something wrong...

$t^4-16=0$

= $(t^2-4)(t^2+4)$

= $(t-2)(t+2)(t+2)(t+2)$

Now should this be the answer $(t-2)(t+2)(t+2)(t+2)$ or should it be this {-2,2}. By solving this I might be able to solve this one $x^4-5x^2+4=0$

Thanks once again guys!!
Hello,

you were asked to find the real solutions of

$t^4-16=0$.......which are $x = -2 ~\vee~x = 2$

With your 2nd problem it would be easier to use a substitution:

If $y = x^2$.... your equation becomes $y^2-5y+4=0$ which can be factored to:

$(y-1)(y-4)=0$....Solve for y and re-substitute these results into the substitution equation and then solve for x.

3. Originally Posted by Morzilla

= $(t^2-4)(t^2+4)$

= $(t-2)(t+2)(t+2)(t+2)$
You should be aware that
$t^2 + 4 \neq (t + 2)^2$

FOIL out the RHS and you will see they can't be equal. There are no real factors of $t^2 + 2$.

(If you like: $t^2 + 4 = (t + 2i)(t - 2i)$.)

-Dan

4. Originally Posted by earboth
Hello,

you were asked to find the real solutions of

$t^4-16=0$.......which are $x = -2 ~\vee~x = 2$

With your 2nd problem it would be easier to use a substitution:

If $y = x^2$.... your equation becomes $y^2-5y+4=0$ which can be factored to:

$(y-1)(y-4)=0$....Solve for y and re-substitute these results into the substitution equation and then solve for x.
Originally Posted by topsquark
You should be aware that
$t^2 + 4 \neq (t + 2)^2$

FOIL out the RHS and you will see they can't be equal. There are no real factors of $t^2 + 2$.

(If you like: $t^2 + 4 = (t + 2i)(t - 2i)$.)

-Dan
ok, (sorry for bothering) but I still don't understand where x=-2 x=2 came from if $t^4-16$ is factored to be $(t^2-4)(t^2+4)$?! or is the factorization of $t^2-4$ the answer to the problem.

Thanks to the both of you for helping out!!

5. ok, (sorry for bothering) but I still don't understand where x=-2 x=2 came from if t^4-16 is factored to be (t^2-4)(t^2+4)?! or is the factorization of t^2-4 the answer to the problem.
We started off with
t^4-16=0
and then factorised to get $(t+2)(t-2)(t^2+4)=0$
This will be true when any of the following are true:
t+2 = 0
t-2 = 0
$
t^2+4 = 0$

because if any of these are 0, then we have 0 times some number which makes 0.

We know that $t^2+4 \not = 0$ for any value of t because $t^2 \geq 0 => t^2+4 \geq 4$

From the first 2 equations we get t = 2 and t = -2.

We started off with
and then factorised to get $(t+2)(t-2)(t^2+4)=0$
This will be true when any of the following are true:
t+2 = 0
t-2 = 0
$
t^2+4 = 0$

because if any of these are 0, then we have 0 times some number which makes 0.

We know that $t^2+4 \not = 0$ for any real value of t because $t^2 \geq 0 => t^2+4 \geq 4$

From the first 2 equations we get t = 2 and t = -2.
Mr F edit in red (sorry, but the original statement is a bug bear of mine )

7. ## Thanks to all

Yes now I understand, the answer is the factorable one $t^2-4$ and not the other $t^2+4$!

I worked out the other problem as well $x^4-5x^2+4=0$ without substitution but factoring, with a lil help of course!

Thanks once again!