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hartnn
 2 years ago
Best ResponseYou've already chosen the best response.1then the two roots of x are: \(\huge{x=\frac{b \pm \sqrt{b^24ac}}{2a}}\)

wio
 2 years ago
Best ResponseYou've already chosen the best response.0Use the quadratic foumula \[x_1 =\frac{ b + \sqrt{b^24ac}} {2a}\] \[x_2 =\frac{ b  \sqrt{b^24ac}} {2a}\]

dddan
 2 years ago
Best ResponseYou've already chosen the best response.0ya but i got 4+ the square root of 24 / 4 does this make sense?

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.1your b=5, how did u get 4+/ ... ?

wio
 2 years ago
Best ResponseYou've already chosen the best response.0You have imaginary roots. Is that what you're worried about?

wio
 2 years ago
Best ResponseYou've already chosen the best response.0Why? Is it a physics problem and you are supposed to have real roots? Because there is nothing wrong with the quadratic formula. Double check your coefficients

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.1u actually have imaginary roots for this..

dddan
 2 years ago
Best ResponseYou've already chosen the best response.0alright can u tell me where to take it from the imaginary roots?

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.1how did u get square root of 24 ?

dddan
 2 years ago
Best ResponseYou've already chosen the best response.0because with the coefficients, its (4)^24(2)(5) , in the square root right?

dddan
 2 years ago
Best ResponseYou've already chosen the best response.0no i i mixed up the coefficents in the original question. i was given a=2;b=4;c=5

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.1ok, \(\huge{x=\frac{4 \pm \sqrt{24}}{4}}=\frac{4 \pm 2\sqrt{6}}{4}=\frac{2\pm i\sqrt{6}}{2}\)

dddan
 2 years ago
Best ResponseYou've already chosen the best response.0alright thank you very much!
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