Mr Asshole Posted August 10, 2013 Share Posted August 10, 2013 Added explanation above. Quote Link to comment Share on other sites More sharing options...
ohhungry Posted August 10, 2013 Share Posted August 10, 2013 4x^2-8x+3 to find roots, use quadratic formula x=(8(+/-)sqrt(8^2-4(4)(3))/8) =(8(+/-)sqrt(16))/8 =(8(+/-)4)/8 =3/2 and 1/2 Set all options equal to 0 and sub in either value of x, whichever statements hold true are correct a)2x+1= 0 (does not work for either value of x) b)4x+3 = 0 (does not work for either value of x) c)2x-3 = 0 (work for x=3/2, therefore this is a factor of 4x^2-8x+3 d)2x+3 = 0 (does not work for either value of x) Therefore 2x-3 is a factor of 4x^2-8x+3, the answer is C Sidenote: You should probably choose harder math problems ahha nope sorry try again, note: ^ means exponent He's actually right... and he did it in a manner much harder then it needed to be. Pfft, trial and error is for nubs Quote Link to comment Share on other sites More sharing options...
mdanhorn Posted August 10, 2013 Share Posted August 10, 2013 4x^2-8x+3 to find roots, use quadratic formula x=(8(+/-)sqrt(8^2-4(4)(3))/8) =(8(+/-)sqrt(16))/8 =(8(+/-)4)/8 =3/2 and 1/2 Set all options equal to 0 and sub in either value of x, whichever statements hold true are correct a)2x+1= 0 (does not work for either value of x) b)4x+3 = 0 (does not work for either value of x) c)2x-3 = 0 (work for x=3/2, therefore this is a factor of 4x^2-8x+3 d)2x+3 = 0 (does not work for either value of x) Therefore 2x-3 is a factor of 4x^2-8x+3, the answer is C Sidenote: You should probably choose harder math problems ahha nope sorry try again, note: ^ means exponent He's actually right... and he did it in a manner much harder then it needed to be. Pfft, trial and error is for nubs There were two plausable choices with that equation, I know I figured it out faster then you did ;) Quote Link to comment Share on other sites More sharing options...
Diamonds Posted August 10, 2013 Author Share Posted August 10, 2013 4x^2-8x+3 to find roots, use quadratic formula x=(8(+/-)sqrt(8^2-4(4)(3))/8) =(8(+/-)sqrt(16))/8 =(8(+/-)4)/8 =3/2 and 1/2 Set all options equal to 0 and sub in either value of x, whichever statements hold true are correct a)2x+1= 0 (does not work for either value of x) b)4x+3 = 0 (does not work for either value of x) c)2x-3 = 0 (work for x=3/2, therefore this is a factor of 4x^2-8x+3 d)2x+3 = 0 (does not work for either value of x) Therefore 2x-3 is a factor of 4x^2-8x+3, the answer is C Sidenote: You should probably choose harder math problems ahha nope sorry try again, note: ^ means exponent Uhh...yeah, my answer's right. unless by 4x^2 you actually meant (4x)^2, in which case there are no real roots. msg me ill answer it rightly, its actually a question that can be done in 2 seconds. Quote Link to comment Share on other sites More sharing options...
mdanhorn Posted August 10, 2013 Share Posted August 10, 2013 4x^2-8x+3 to find roots, use quadratic formula x=(8(+/-)sqrt(8^2-4(4)(3))/8) =(8(+/-)sqrt(16))/8 =(8(+/-)4)/8 =3/2 and 1/2 Set all options equal to 0 and sub in either value of x, whichever statements hold true are correct a)2x+1= 0 (does not work for either value of x) b)4x+3 = 0 (does not work for either value of x) c)2x-3 = 0 (work for x=3/2, therefore this is a factor of 4x^2-8x+3 d)2x+3 = 0 (does not work for either value of x) Therefore 2x-3 is a factor of 4x^2-8x+3, the answer is C Sidenote: You should probably choose harder math problems ahha nope sorry try again, note: ^ means exponent Uhh...yeah, my answer's right. unless by 4x^2 you actually meant (4x)^2, in which case there are no real roots. msg me ill answer it rightly, its actually a question that can be done in 2 seconds. Uhh.. what level of math are you in? As he displayed (at minimum) lower level calculus understanding and implentation, and his answer is correct. So I don't understand how you can suggest his answer is wrong, when it's quite right given the equation provided to us. Quote Link to comment Share on other sites More sharing options...
Diamonds Posted August 10, 2013 Author Share Posted August 10, 2013 4x^2-8x+3 to find roots, use quadratic formula x=(8(+/-)sqrt(8^2-4(4)(3))/8) =(8(+/-)sqrt(16))/8 =(8(+/-)4)/8 =3/2 and 1/2 Set all options equal to 0 and sub in either value of x, whichever statements hold true are correct a)2x+1= 0 (does not work for either value of x) b)4x+3 = 0 (does not work for either value of x) c)2x-3 = 0 (work for x=3/2, therefore this is a factor of 4x^2-8x+3 d)2x+3 = 0 (does not work for either value of x) Therefore 2x-3 is a factor of 4x^2-8x+3, the answer is C Sidenote: You should probably choose harder math problems ahha nope sorry try again, note: ^ means exponent Uhh...yeah, my answer's right. unless by 4x^2 you actually meant (4x)^2, in which case there are no real roots. msg me ill answer it rightly, its actually a question that can be done in 2 seconds. Uhh.. what level of math are you in? As he displayed (at minimum) lower level calculus understanding and implentation, and his answer is correct. So I don't understand how you can suggest his answer is wrong, when it's quite right given the equation provided to us. the answer i have is based on a school textbook, maybe he is trying to figure out something wrong, or has made a mistake, the answer is 4x + 3 so someone try to explain why. Quote Link to comment Share on other sites More sharing options...
ohhungry Posted August 10, 2013 Share Posted August 10, 2013 There were two plausable choices with that equation, I know I figured it out faster then you did Nah, I knew the answer already, but I wanted the longest description cause apparently thats the contest or something :P Quote Link to comment Share on other sites More sharing options...
ohhungry Posted August 10, 2013 Share Posted August 10, 2013 the answer i have is based on a school textbook, maybe he is trying to figure out something wrong, or has made a mistake, the answer is 4x + 3 so someone try to explain why. Maybe we get a prize for telling you how you are wrong? Obviously your textbook is wrong or you're just trolling everybody... if 4x+3 was a factor, and we put it in the form (x-s)(x-t) you would then have (4x+3)(x-t)=4x^2-8x+3 Now if you expand out those brackets, the only way to have 3 on both sides is if t= -1, and if t= -1, there will -x on the left side and -8x on the right side, and the equation would obviously not be true for all values of x, hence 4x+3 is not a factor. Quote Link to comment Share on other sites More sharing options...
mdanhorn Posted August 10, 2013 Share Posted August 10, 2013 Indeed the textbook is wrong, or you wrote out the wrong equation.. As it is impossible to get: 4x^2-8x+3 from 4x+3 In order to get a negative middle term, the 2nd portion of the other factor would have to be negative, which would in turn make 3 in the original equation negative.. Quote Link to comment Share on other sites More sharing options...
