ohhungry Posted August 10, 2013 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
mdanhorn Posted August 10, 2013 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 ;)
Diamonds Posted August 10, 2013 Author 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.
mdanhorn Posted August 10, 2013 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.
Diamonds Posted August 10, 2013 Author 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.
ohhungry Posted August 10, 2013 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
ohhungry Posted August 10, 2013 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.
mdanhorn Posted August 10, 2013 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..
