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

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sabika13 Group TitleBest ResponseYou've already chosen the best response.0
quadratic formula and factoring
 one year ago

calculusfunctions Group TitleBest ResponseYou've already chosen the best response.0
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.
 one year ago

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

chicagochica5 Group TitleBest ResponseYou've already chosen the best response.0
Which is a better way the quadratic formula or the square and why
 one year ago

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

chicagochica5 Group TitleBest 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
 one year ago

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

dylanhouse Group TitleBest ResponseYou've already chosen the best response.0
It has the \[x^2\] term.
 one year ago

CliffSedge Group TitleBest ResponseYou've already chosen the best response.1
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 plugandchug in a formula.
 one year ago

CliffSedge Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago

chicagochica5 Group TitleBest 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??
 one year ago

chicagochica5 Group TitleBest ResponseYou've already chosen the best response.0
@satellite73
 one year ago

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

CliffSedge Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago

chicagochica5 Group TitleBest ResponseYou've already chosen the best response.0
theequation is not factorable @CliffSedge ?
 one year ago

CliffSedge Group TitleBest ResponseYou've already chosen the best response.1
Like calculusfunctions said, not with integers.
 one year ago

CliffSedge Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago
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