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Thread: Solving quadratic equations by factoring

  1. #1
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    Solving quadratic equations by factoring

    Using the "zero property" $\displaystyle ax^2+bx+c=0$

    $\displaystyle 4x^2=60-x$

    move the $\displaystyle 60-x$ over to the left...

    $\displaystyle 4x^2-60+x=0$

    Factor out...

    ok now here is where I have my problem. As seeing that the Eq has a negative and a positive the signs would be the same when factoring out no?!

    $\displaystyle (4x-15)(x-4)$

    $\displaystyle 4x-15=0$ and $\displaystyle x-4=0$

    $\displaystyle 4x=+15$ and $\displaystyle x=4$

    Thus...

    $\displaystyle x=\frac{15}{4}$ and $\displaystyle x=4$

    Yes?!
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  2. #2
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    your factoring was incorrect.

    $\displaystyle 4x^2-60+x = (4x- 15)(x+4)$

    remember you can always check if your factorisation is correct by expanding your factors.
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  3. #3
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    Quote Originally Posted by bobak View Post
    your factoring was incorrect.

    $\displaystyle 4x^2-60+x = (4x- 15)(x+4)$

    remember you can always check if your factorisation is correct by expanding your factors.
    .....ok how do I expand the factors, that's something I have not heard of!?



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  4. #4
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    you just multiply out the factors.

    so $\displaystyle (4x- 15)(x+4) = 4x \cdot x + 4x \cdot 4 -15 \cdot x - 15 \cdot 4$
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  5. #5
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    Quote Originally Posted by bobak View Post
    you just multiply out the factors.

    so $\displaystyle (4x- 15)(x+4) = 4x \cdot x + 4x \cdot 4 -15 \cdot x - 15 \cdot 4$
    ahhhhhh yes yes!! In the original problem I did not re-arrange the equation so that it would be in standard form!

    $\displaystyle 4x^2+x-60$

    Now it makes sense!!

    thanks Bobak
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