You can always find solutions to a quadratic equation (real or complex) if you use the quadratic formula:
x = [-b ± √(b² - 4ac)]/2a
In your problem:
a = 9 (coefficient of the x² term)
b = -14 (coefficient of the x term)
c = -4 (constant term)
Plugging in a, b, and c into the quadratic formula:
x = [-(-14) ± √((-14)² - 4(9)(-4))]/2(9)
=[14 ± √340]/18
So you have two answers:
x = [14 + √340]/18 = 1.802
OR
x = [14 - √340]/18 = -0.247
2007-01-09 18:17:37
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answer #1
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answered by alsh 3
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X = 1
2007-01-10 02:49:57
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answer #2
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answered by Adios 7
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to solve quadratic equation that the first term ( x2 )have a coefficient, you multiply the coefficient with the constant term -4.
determine two number that when you multiply will give the product of the first and last term and when added will give the sum of the middle term.
in this case the equation have no solution.
1.because the sum of all # can not give -14x.
2.maybe there is a mistake with the middle term -14x.
check the number again.i think it should be -16x
the answer is-- not factorable or no solution.
i tried but I'm sure there is no solution to this,good luck
2007-01-10 11:03:25
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answer #3
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answered by bright 247 2
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9x^2 - 14x - 4 = 0
We use the quadratic formula.
x = [-b +/- sqrt(b^2 - 4ac)] / [2a]
x = [14 +/- sqrt(340)] / 18
x = [14 +/- 4sqrt(85)] / 18
Since all of our numeric values are even,
x = [7 +/- 2sqrt(85)]/9
So our two answers are:
x = {[7 + 2sqrt(85)]/9 , [7 - 2sqrt(85)]/9}
2007-01-10 02:21:05
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answer #4
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answered by Puggy 7
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x = [14 +/- sqrt(196 +144)]/18
= [14 +/- sqrt(340)]/18 = [14 +/- 18.44]/18
x = -0.2466 or 1.8022 (approximately)
2007-01-10 02:16:07
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answer #5
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answered by heartsensei 4
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x= 1.802171606
x= -0.24661605
2007-01-10 09:09:41
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answer #6
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answered by libra 1
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1.801 , -0.246
2007-01-10 02:15:09
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answer #7
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answered by catfan 1
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