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chicagochica5
 3 years ago
What method(s) would you choose to solve the equation? Explain your reasoning.
2x^2 +4x  3 = 0
chicagochica5
 3 years ago
What method(s) would you choose to solve the equation? Explain your reasoning. 2x^2 +4x  3 = 0

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sabika13
 3 years ago
Best ResponseYou've already chosen the best response.0quadratic formula and factoring

calculusfunctions
 3 years ago
Best ResponseYou've already chosen the best response.0Factoring 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.

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.0i 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 \[(x+1)^2=\frac{3}{2}+1=\frac{5}{2}\]

chicagochica5
 3 years ago
Best ResponseYou've already chosen the best response.0Which is a better way the quadratic formula or the square and why

dylanhouse
 3 years ago
Best ResponseYou've already chosen the best response.0Depends which you prefer. Quadratic is easier to some. Completing the square just requires you be a little more tedious :)

chicagochica5
 3 years ago
Best ResponseYou've already chosen the best response.0@dylanhouse, i do not think i need to solve.I think i just need to say which formala would i use and why

dylanhouse
 3 years ago
Best ResponseYou've already chosen the best response.0Well, you would use the quadratic formula \[x=\frac{ b \pm \sqrt{b^24ac} }{ 2a }\] simply because it is a quadratic equation!

dylanhouse
 3 years ago
Best ResponseYou've already chosen the best response.0It has the \[x^2\] term.

CliffSedge
 3 years ago
Best ResponseYou've already chosen the best response.1The 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 plugandchug in a formula.

CliffSedge
 3 years ago
Best ResponseYou've already chosen the best response.1I'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.

chicagochica5
 3 years ago
Best ResponseYou've already chosen the best response.0@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??

CliffSedge
 3 years ago
Best ResponseYou've already chosen the best response.1Sounds good to me, though you should probably also include that you chose that because it is not factorable.

CliffSedge
 3 years ago
Best ResponseYou've already chosen the best response.1Though 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.

chicagochica5
 3 years ago
Best ResponseYou've already chosen the best response.0theequation is not factorable @CliffSedge ?

CliffSedge
 3 years ago
Best ResponseYou've already chosen the best response.1Like calculusfunctions said, not with integers.

CliffSedge
 3 years ago
Best ResponseYou've already chosen the best response.1To 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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