2006-11-12 13:59:57 · 5 answers · asked by khalid a 1 in Science & Mathematics ➔ Mathematics
3x^2-16x+17=0
x = [-(-16) +- sqrt( (-16)^2 - 4(3)(17))]/[2(3)]
x = [(16) +- sqrt(256 - 204)]/[6]
x = [(16) +- sqrt(52)]/[6]
x = [(16) + sqrt(52)]/[6] or [(16) - sqrt(52)]/[6]
x = [(16) + 2*sqrt(13)]/[6] or [(16) - 2*sqrt(13)]/[6]
x = [8 + sqrt(13)]/3 or [8 - sqrt(13)]/3
2006-11-12 14:11:04 · answer #1 · answered by wheezer_april_4th_1966 7 · 3⤊ 0⤋
this is not a simple quadratic--- for this you have to use this equation
(-b+/- rad(b^2 - 4ac))/ 2a
a= 3 b= -16 c= 17
(16+/- rad((-16^2) - 4(3)*(17))))/ 2(3)
at the end when u solve u get
(16 +/- 7.2)/6
so now u first add and divide by 6, and then you subtract and divide by six and ur answers are,
3.87 and 1.47
2006-11-12 22:08:33 · answer #2 · answered by aplpie 3 · 3⤊ 0⤋
Use the quadratic formula:
x=-b屉b^2-4ac/2a
a=3, b=-16, c=17
I will let you figure out what to do next since you need to learn how to do this on your own!
2006-11-12 22:07:20 · answer #3 · answered by Anonymous · 1⤊ 1⤋
general quadratic formula
ax^2+bx+c;a=3;b=-16;c=17.sqrt=square root
[-b+/- sqrt((-b)^2-4ac)]/2a
[16+/-sqrt(256-204)]/6
[16+sqrt(52)]/6
(16+7.211)/6=3.8685
Or [16-sqrt(52)]/6
(16-7.211)/6=1.4648
2006-11-12 22:09:59 · answer #4 · answered by Phy A 5 · 3⤊ 0⤋
Do the quadratic formula.
2006-11-12 22:02:52 · answer #5 · answered by Habester 3 · 2⤊ 1⤋