the formula is -b +or- the square root of b^2-4ac all over 2a
a=1
b=-2
c=12
(2+,-√4-48)/2
the answer would be (1+,-i√11)/2
2007-05-23 10:52:52
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answer #1
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answered by vols_fan 2
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Quadratic equations are of the general form:
ax^2+bx+c=0 where a<>0.
In your case x^2-2x+12=0 you have:
a=1
b=-2
c=12
We culculate:
D=b^2-4ac=(-2)^2-4*1*12=4-48=-44<0
So, your equation has not any real roots, but the complex:
x1=(2+i*sqrt(44))/2 = 1 +i*sqrt(11) and
x2=(2-i*sqrt(44))/2 = 1 -i*sqrt(11)
2007-05-23 17:59:35
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answer #2
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answered by Anonymous
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the quadratic formula is neg B plus or minus the square root of B squared minus 4AC all over 2 A...
so you need the plug them all in
-(-2)+/- sqr root((-2)^2-4(1)(12)) all divided by 2(1)
do whats in the sqr root first so you get 4-48=-44
since what you get is negative it means you get an imaginary number (i), which means it doesnt go through the x axis when graphed
now to solve it, you do the rest of the problem which is (2+/- sqr root(-44)) / 2
the sqr root of -44 is simplified to 2i((sqr rt)-11)
that divided by 2 is i(sq rt -11) and 2divided by 2 is one so you end up with x=1+/-i(sq rt-11)
sorry that its so confusing =(
2007-05-23 18:14:14
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answer #3
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answered by Larvay 1
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x² - 2x+12=0
ax² + bx + c = 0
x = (-b +- â(b² - 4ac))/2a
So in your equation
x = (2 +- â(4 - 48))/2
x = (2 +- â(-44))/2
Take 4 out of the radical and it's root is 2
x = (2 +- 2â(-11))/2
Take the root of -1 outside of the radical, it is i
x = (2 +- 2iâ(11))/2
Divide numerator and denominator by 2
x = 1 + iâ(11), x = 1 - iâ(11)
.
2007-05-23 17:55:00
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answer #4
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answered by Robert L 7
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-b±âb² - 4ac 2±â-2² - 4(1)(12)
----------------- -----------------------
2a 2(1)
^ ^ ^
Quadratic your equation plugged in
formula
the x² term = a
the x term = b
the # term = c
Just plug it in then use order of operations to simplify
by the way the ( ) means to multiply
2007-05-23 17:54:04
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answer #5
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answered by coolhomie9 1
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