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6 Comments Already

mygif
Puzzling Said,
February 5th, 2011 @5:22 pm  

You have a polynomial of the form:
ax² + bx + c = 0

In your example:
a = 1/3
b = -2
c = 9

Now plug those values into the quadratic formula:
…… -b ± ?(b² – 4ac)
x = ———————-
…………… 2a

…… -(-2) ± ?((-2)² – 4(1/3)(9))
x = ———————————-
…………… 2(1/3)

…… 2 ± ?(4 – 12)
x = ——————-
……….. (2/3)

…… 2 ± ?(-8)
x = ————–
……….. (2/3)

Invert the denominator and multiply:
x = [ 2 ± ?(-8) ] * (3/2)

Distribute 3/2 through:
x = 2 * (3/2) ± ?(-8) * (3/2)

Simplify:
x = 3 ± ?(-8) * (3/2)

Now the complication is the negative square root, so we will have an imaginary number.
x = 3 ± i?8 * (3/2)
x = 3 ± i?(4 * 2) * (3/2)
x = 3 ± i?4 * ?2 * (3/2)
x = 3 ± 2i * ?2 * (3/2)
x = 3 ± 3i * ?2
x = 3 ± (3?2)i

Your two answers are:
x = 3 + (3?2)i
or
x = 3 – (3?2)i

mygif
eltonjohn_111 Said,
February 5th, 2011 @6:06 pm  

first get rid of the fraction so multiply everything by three so you get x^2-6x+27 and then use the -b plus or minus the square root of b^2-4ac all divided by 2a and when u find the plus or minuses u will have your in this case two zeros

mygif
Alec B Said,
February 5th, 2011 @6:35 pm  

The formula is (-b+/-sqrt(b^2-4ac))/2a, where ‘a’ is the coefficient of x^2, ‘b’ the coefficient of x and ‘c’ the constant term (and ‘sqrt’ means ‘square root).

In your case a=1/3, b=-2 and c=9.

so x=(2+/-sqrt(4-4/3))/(2/3)

=3+/-3*sqrt(1-1/3)

=3+sqrt(6) OR 3-sqrt(6)

Those two values are the zeroes of the function.

HTH

mygif
aBitRusty Said,
February 5th, 2011 @7:16 pm  

for a polynomial
ax^2 + bx + c = 0

the quadratic formula will give the roots:
the formula:

x1,x2 = (-b/2a) +/- (1/2a)*sqrt(b^2 – 4ac)
+/- meaning two solutions, one + .. and one -

so, here
y = 1/3x^2 – 2x + 9
y= 0
0 = 1/3x^2 – 2x + 9
a = 1/3 , b= -2, c= 9

fill all in in formula:
x1,x2 = (2/(2/3)) +/- (1/2(1/3))*sqrt(2^2 – 4*1/3*9)
x1,x2 = 3 +/- (3/2)*sqrt(4 – 12)
x1,x2= 3+/- (2/2)* sqrt(-8)

now, not sure if you covered imaginairy numbers yet.
if not, the sqrt(-8) means you’ve no solutions
(the discriminant part of the quadr. formula ‘b^-4ac’
less than 0)

otherwise:
x1,x2 = 3+/-(3/2)*sqrt(i^2*4*2)
x1,x2=3+/-3i*sqrt(2)

so x = 3+3isqrt(2) or x = 3 = 3-3i*sqrt(2)

hope that helped.

mygif
D.W. Said,
February 5th, 2011 @7:25 pm  

The zeroes are the values of x that make y=0.
(1/3)x² – 2x + 9 = 0
Use the quadratic formula to solve
   ax²+bx+c=0
where
   a= (1/3)
   b=-2
   c=9

x = [-b ± ?(b²-4ac)] / (2a)
   = [-(-2) ± ?((-2)² - 4((1/3))(9))] / 2((1/3))
   = [2 ± ?(-8)] / (2/3)
   = [2 ± ?(-1)?(2³)] /(2/3)
   = [2 ± 2i?2] / (2/3)
   = 3 ± 3i?2

x? = 3 + 4.243i
x? = 3 – 4.243i

mygif
como Said,
February 5th, 2011 @8:23 pm  

x² – 6x + 27 = 0
x = [ 6 ± ? (36 - 108 ) ] / 2
x = [ 6 ± ? (- 72 ) ] / 2
x = [ 6 ± i 6? (2 ) ] / 2
x = 3 ± i 3 ? (2 )
x = 3 (1 ± i ? 2 )

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