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Thread: Quadratic in form

  1. #1
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    Quadratic in form

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

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

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

    $\displaystyle t^4-16=0$

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

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

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

    Thanks once again guys!!
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  2. #2
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    Quote Originally Posted by Morzilla View Post
    I know how to factor this $\displaystyle t^2-16=0$ $\displaystyle (t-4)(t+4)$ but today I am asked to find the real solutions for equations that are quadratic in form....

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

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

    $\displaystyle t^4-16=0$

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

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

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

    Thanks once again guys!!
    Hello,

    you were asked to find the real solutions of

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

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

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

    $\displaystyle (y-1)(y-4)=0$....Solve for y and re-substitute these results into the substitution equation and then solve for x.
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Morzilla View Post

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

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

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

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

    -Dan
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  4. #4
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    Quote Originally Posted by earboth View Post
    Hello,

    you were asked to find the real solutions of

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

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

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

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

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

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

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

    Thanks to the both of you for helping out!!

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  5. #5
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    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 $\displaystyle (t+2)(t-2)(t^2+4)=0$
    This will be true when any of the following are true:
    t+2 = 0
    t-2 = 0
    $\displaystyle
    t^2+4 = 0$

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

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

    From the first 2 equations we get t = 2 and t = -2.
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  6. #6
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    Quote Originally Posted by badgerigar View Post
    We started off with
    and then factorised to get $\displaystyle (t+2)(t-2)(t^2+4)=0$
    This will be true when any of the following are true:
    t+2 = 0
    t-2 = 0
    $\displaystyle
    t^2+4 = 0$

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

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

    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 )
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  7. #7
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    Thanks to all

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

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

    Thanks once again!

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