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What method(s) would you choose to solve the equation? Explain your reasoning. 2x^2 +4x - 3 = 0

Mathematics
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quadratic formula and factoring
Factoring over integers is not possible, thus solving by the quadratic formula would be the most efficient method. You could also solve by the method of completing the square.
i would complete the square because after \[2x^2+4x=3\] and \[x^2+2x=\frac{3}{2}\] the coefficient of the middle term is even so you can go right to \[(x+1)^2=\frac{3}{2}+1=\frac{5}{2}\]

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Other answers:

Which is a better way the quadratic formula or the square and why
Depends which you prefer. Quadratic is easier to some. Completing the square just requires you be a little more tedious :)
@dylanhouse, i do not think i need to solve.I think i just need to say which formala would i use and why
Well, you would use the quadratic formula \[x=\frac{ -b \pm \sqrt{b^2-4ac} }{ 2a }\] simply because it is a quadratic equation!
It has the \[x^2\] term.
The quadratic formula is the result of completing the square. They are equivalent. You choose one over the other depending on the coefficients of the terms. Some numbers are easy to play with algebraically (small integers, etc.), others (big, clunky decimals) would probably be easier to just plug-and-chug in a formula.
I'm with satellite on this one. I'd complete the square. The numbers are small and friendly, and it only takes a couple steps to move things around.
@CliffSedge,(my answer) My method to solve the equation would be completing the sqaure... and I can just put my reasoning because the numbers are small and it only takes a couple steps??
Sounds good to me, though you should probably also include that you chose that because it is not factorable.
Though sometimes I have a quadratic equation that is factorable, and I use the quadratic formula anyway because I'm feeling lazy. "Feeling lazy" is a perfectly acceptable reason sometimes - as long as you still get the right answer with whatever method you choose.
theequation is not factorable @CliffSedge ?
Like calculusfunctions said, not with integers.
To see, take the leading coefficient, 2, and the constant, -3 from 2x^2 +4x - 3 and multiply them and get -6 There are no factors of -6 that add to make 4.

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