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I am suppose to find all the solutions to this problem, I think some theorem states that there can only be as many solutions to the problem as the highest degree. I know that calculus reinforces this so I know that

$2x^2 + 4x + 1 = 0$

Can have at most two solutions. In calculus this is proven by the derivative being zero at only somewhere. I can't remember and it isn't important yet.

Anyways I have no idea what to do with this problem. I don't think I can factor it conventionally because of the 2 coefficient so what is the method at this point? I tried guessing and it didn't work at all for -2 - 3.

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    Then you need to commit to memory either an alternative solution method, the means to re-derive it when necessary, or where to look it up when necessary.2012-12-04

5 Answers 5

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You should definitely commit the formula to memory. Here's a cool video to the tune of row-row-row-your-boat, but there are plenty more out there. http://www.youtube.com/watch?v=HRcj9slciqM. Once you get on youtube, search around as there are lots of great videos in there that show you how to use the formula.

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$ax^2 + bx + c = 0, \text{ roots?}$
This is where the quadratic formula comes in handy (you should memorize this!):


$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$


The discriminant d of the quadratic formula is the term $d= (b^2 - 4ac)$. When $b^2\geq 4ac$, you have two real roots.

At the very least, try to memorize how to compute the discriminant. That will allow you to easily determine whether or not a quadratic equation has real roots.

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If you don't want to use the formula, you can do the following.(Actually this is how the formula is derived though. I used to use this method when I didn't memorize the formula.)

$2x^2+4x+1=0$

$x^2+2x+1/2=0$

$(x+1)^2-1/2=0$

$(x+1)^2=1/2$

$x+1=+1/\sqrt 2$ or $-1/\sqrt 2$

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Use the Quadratic Formula. One cannot expect all quadratic polynomials to factor "nicely." The Quadratic Formula needs to become a completely standard tool in your arsenal.

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    The reason that in the past you could guess roots is that the problem-setter chose the coefficients $a$, $b$, and $c$ of $ax^2+bx+c$ so that the roots were easily guessable: either small integers, or, sometimes, rationals with simple denominators. In many of the problems you will meet, the roots are not guessable. If one chooses $a$, $b$, $c$ at random, most of the time the roots are not guessable.2012-12-03
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The quadratic formula works for all quadratic equations. For any equation [in the form] $ax^2 + bx + c = 0$, this is true:$x = {-b \pm \sqrt{b^2 - 4ac} \over 2a}$