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2007-08-27 04:29:57
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answer #1
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answered by Anonymous
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There are 3 ways to solve quadratic equations.
Some will factor. For example,
2x² + 5x - 3 = 0
(2x - 1)(x + 3) = 0
then set each factor to 0 and solve:
2x - 1 = 0
2x = 1
x = 1/2
x + 3 = 0
x = -3
Some are easy to solve by completing the square. Best if number in front of x² is 1 and number in front of x is even:
x² - 6x + 2 = 0
x² - 6x = -2
x² - 6x + 9 = -2 + 9
the 9 is (-6/2)²
(x - 3)² = 7
x - 3 = ±√7
x = 3 ± √7
but all quadratics can be solved using the quadratic formula: if ax² + bx + c = 0, then
x = -b/(2a) ± √(b² - 4ac) / 2a
3x² + 4x - 2 = 0
x = -4/6 ± √(16 - 4(-6)) / 6
x = -2/3 ± √40 / 6
x = -2/3 ± 2√10 / 6
x = -2/3 ± √10 / 3
it's actually more work to simplify the answer than to fill in the formula.
2007-08-27 04:40:36
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answer #2
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answered by Philo 7
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Textbooks often say there are three ways to solve a quadratic equation, namely, factoring, completing the square, and by the quadratic formula. The last method is easiest to use. The formula is this: to solve ax^2 + bx + c , with a not 0, take x = (-b +/- sqrt(b^2 - 4ac))/2a .
Furthermore, the discriminant is D = b^2 - 4ac. If D > 0, there are two real and distinct roots; if D = 0, there are two real and equal roots; and if D < 0, there are two complex roots which are conjugates.
2007-08-27 04:43:30
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answer #3
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answered by Tony 7
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by the way....a way to remember the quadratic formula is the following story : A negative boy (-b) couldn't decide (+ or -) whether or not to go to a radical party (sq. root of). The boy was squared and he missed out on 4 awesome chics (b square minus 4ac). The party was all over at 2am (all divided by 2a).
2007-08-27 04:55:43
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answer #4
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answered by kat79015 2
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2016-10-17 02:46:15
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answer #5
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answered by ? 4
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http://www.themathpage.com/alg/algebra.htm
look check this site and search for what u want is a great site and explains how to do anything : )
2007-08-27 04:33:29
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answer #6
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answered by The Answer 3
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